Sau đây là Hướng dẫn và lời giải chi tiết các bài tập của Bài 4 – Chương 1, trong sách giáo khoa môn Toán lớp 7 – tập 1, thuộc bộ sách Chân trời sáng tạo.
Thực hành 1 (Trang 22 / Toán 7 – tập 1 / Chân trời sáng tạo) Cho biểu thức:
$$A = \left(7 – \frac{2}{5} + \frac{1}{3}\right) – \left(6 – \frac{4}{3} + \frac{6}{5}\right) – \left(2 – \frac{8}{5} + \frac{5}{3}\right).$$
Hãy tính giá trị của $A$ bằng cách bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.
Giải
$$A = \left(7 – \frac{2}{5} + \frac{1}{3}\right) – \left(6 – \frac{4}{3} + \frac{6}{5}\right) – \left(2 – \frac{8}{5} + \frac{5}{3}\right)$$
$$= 7- \frac{2}{5} + \frac{1}{3} – 6 + \frac{4}{3} – \frac{6}{5} – 2 + \frac{8}{5} – \frac{5}{3}$$
$$= 7 – 6 – 2 – \frac{2}{5} – \frac{6}{5} + \frac{8}{5} + \frac{1}{3} + \frac{4}{3} – \frac{5}{3}$$
$$= (7 – 6 – 2) – \left(\frac{2}{5} + \frac{6}{5} – \frac{8}{5}\right) + \left(\frac{1}{3} + \frac{4}{3} – \frac{5}{3}\right)$$
$$= (-1) – 0 + 0$$
$$= -1$$
Thực hành 2 (Trang 23 / Toán 7 – tập 1 / Chân trời sáng tạo) Tìm $x,$ biết:
$$\mathbf{a)}\; x + \frac{1}{2} = -\frac{1}{3};$$
$$\mathbf{b)}\; \left(-\frac{2}{7}\right) + x = -\frac{1}{4}.$$
Giải
$$\mathbf{a)}\; x + \frac{1}{2} = -\frac{1}{3}$$
$$x = -\frac{1}{3} – \frac{1}{2} = \frac{-2}{6} – \frac{3}{6} = \frac{-5}{6}.$$
$$\mathbf{b)}\; \left(-\frac{2}{7}\right) + x = -\frac{1}{4}$$
$$x = -\frac{1}{4} + \frac{2}{7} = \frac{-7}{28} + \frac{8}{28} = \frac{1}{28}.$$
Thực hành 3 (Trang 24 / Toán 7 – tập 1 / Chân trời sáng tạo) Tính:
$$\mathbf{a)}\; 1\frac{1}{2} + \frac{1}{5}\cdot \left[\left(-2\frac{5}{6} \right)+ \frac{1}{3}\right];$$
$$\mathbf{b)}\; \frac{1}{3}\cdot \left(\frac{2}{5} -\frac{1}{2}\right): \left(\frac{1}{6} – \frac{1}{5}\right)^2.$$
Giải
$$\mathbf{a)}\; 1\frac{1}{2} + \frac{1}{5}\cdot \left[\left(-2\frac{5}{6} \right)+ \frac{1}{3}\right]$$
$$= \frac{3}{2} + \frac{1}{5} \cdot \left[\frac{-17}{6} + \frac{1}{3}\right]$$
$$= \frac{3}{2} + \frac{1}{5} \cdot \left[\frac{-17}{6} + \frac{2}{6}\right]$$
$$= \frac{3}{2} + \frac{1}{5} \cdot \frac{-15}{6}$$
$$= \frac{3}{2} + \frac{-1}{2}$$
$$= \frac{2}{2}$$
$$= 1.$$
$$\mathbf{b)}\; \frac{1}{3}\cdot \left(\frac{2}{5} -\frac{1}{2}\right): \left(\frac{1}{6} – \frac{1}{5}\right)^2$$
$$= \frac{1}{3} \cdot \left(\frac{4}{10} – \frac{5}{10}\right) : \left(\frac{5}{30} – \frac{6}{30}\right)^2$$
$$= \frac{1}{3} \cdot \frac{-1}{10} : \left(\frac{-1}{30}\right)^2$$
$$= \frac{-1}{30} : \left(\frac{-1}{30}\right)^2$$
$$= \frac{-1}{30} : \frac{1}{30^2}$$
$$= \frac{-1}{30} \cdot 30^2$$
$$= -30.$$
Lưu ý
Trong câu b), đến đoạn: $\frac{-1}{30} : \left(\frac{-1}{30}\right)^2,$ ta có thể tính như sau:
Đặt $m = \frac{-1}{30}$ thì ta có:
$$\frac{-1}{30} : \left(\frac{-1}{30}\right)^2 = m : (m)^2$$
$$= \frac{m}{m^2} = \frac{1}{m} = 1 : m$$
$$= 1 : \frac{-1}{30} = -30$$
Bài tập 1 (Trang 24 / Toán 7 – tập 1 / Chân trời sáng tạo) Bỏ dấu ngoặc rồi tính:
$$\mathbf{a)}\; \left(\frac{-3}{7}\right)+\left(\frac{5}{6} – \frac{4}{7}\right);$$
$$\mathbf{b)}\; \frac{3}{5} – \left(\frac{2}{3} + \frac{1}{5}\right);$$
$$\mathbf{c)}\; \left[\left(\frac{-1}{3}\right)+1\right] -\left(\frac{2}{3} – \frac{1}{5}\right);$$
$$\mathbf{d)}\; 1\frac{1}{3} + \left(\frac{2}{3} – \frac{3}{4}\right) – \left(0,8 + 1\frac{1}{5}\right).$$
Giải
$$\mathbf{a)}\; \left(\frac{-3}{7}\right)+\left(\frac{5}{6} – \frac{4}{7}\right)$$
$$= \frac{-3}{7} + \frac{5}{6} – \frac{4}{7}$$
$$= \frac{-3}{7} – \frac{4}{7} + \frac{5}{6}$$
$$= \left(\frac{-3}{7} – \frac{4}{7}\right) + \frac{5}{6}$$
$$= \frac{-7}{7} + \frac{5}{6}$$
$$= -1 + \frac{5}{6}$$
$$= \frac{-6}{6} + \frac{5}{6}$$
$$= \frac{-1}{6}.$$
$$\mathbf{b)}\; \frac{3}{5} – \left(\frac{2}{3} + \frac{1}{5}\right)$$
$$= \frac{3}{5} – \frac{2}{3} – \frac{1}{5}$$
$$= \frac{3}{5} – \frac{1}{5} – \frac{2}{3}$$
$$= \left(\frac{3}{5} – \frac{1}{5}\right) – \frac{2}{3}$$
$$= \frac{2}{5} – \frac{2}{3}$$
$$= \frac{6}{15} – \frac{10}{15}$$
$$= \frac{-4}{15}.$$
$$\mathbf{c)}\; \left[\left(\frac{-1}{3}\right)+1\right] -\left(\frac{2}{3} – \frac{1}{5}\right)$$
$$= \left[\frac{-1}{3} + 1\right] – \frac{2}{3} + \frac{1}{5}$$
$$= \frac{-1}{3} + 1 – \frac{2}{3} + \frac{1}{5}$$
$$= \frac{-1}{3} – \frac{2}{3} + 1 + \frac{1}{5}$$
$$= \left(\frac{-1}{3} – \frac{2}{3}\right) + 1 + \frac{1}{5}$$
$$= \frac{-3}{3} + 1 + \frac{1}{5}$$
$$= -1 + 1 + \frac{1}{5}$$
$$= \frac{1}{5}.$$
$$\mathbf{d)}\; 1\frac{1}{3} + \left(\frac{2}{3} – \frac{3}{4}\right) – \left(0,8 + 1\frac{1}{5}\right)$$
$$= 1\frac{1}{3} + \frac{2}{3} – \frac{3}{4} – 0,8 – 1\frac{1}{5}$$
$$= \frac{4}{3} + \frac{2}{3} – \frac{3}{4} – \frac{4}{5} – \frac{6}{5}$$
$$= \left(\frac{4}{3} + \frac{2}{3}\right) – \frac{3}{4} – \left(\frac{4}{5} + \frac{6}{5}\right)$$
$$= \frac{6}{3} – \frac{3}{4} – \frac{10}{5}$$
$$= 2 – \frac{3}{4} – 2$$
$$= 2 – 2 – \frac{3}{4}$$
$$= -\frac{3}{4}.$$
Bài tập 2 (Trang 25 / Toán 7 – tập 1 / Chân trời sáng tạo) Tính:
$$\mathbf{a)}\; \left(\frac{3}{4}:1\frac{1}{2}\right)- \left(\frac{5}{6} : \frac{1}{3}\right);$$
$$\mathbf{b)}\; \left[\left(\frac{-1}{5}\right):\frac{1}{10}\right]- \frac{5}{7}\cdot \left(\frac{2}{3} – \frac{1}{5}\right);$$
$$\mathbf{c)}\; (-0,4)+ 2\frac{2}{5}\cdot \left[\left(\frac{-2}{3}\right)+\frac{1}{2}\right]^2;$$
$$\mathbf{d)}\; \left\{ \left[ \left(\frac{1}{25}- 0,6\right)^2 : \frac{49}{125}\right]\cdot \frac{5}{6} \right\} – \left[\left(\frac{-1}{3}\right)+\frac{1}{2}\right].$$
Giải
$$\mathbf{a)}\; \left(\frac{3}{4}:1\frac{1}{2}\right)- \left(\frac{5}{6} : \frac{1}{3}\right)$$
$$= \left(\frac{3}{4} : \frac{3}{2}\right) – \left(\frac{5}{6} \cdot 3\right)$$
$$= \left(\frac{3}{4} \cdot \frac{2}{3} \right) – \frac{5}{2}$$
$$= \frac{1}{2} – \frac{5}{2}$$
$$= \frac{-4}{2} = -2.$$
$$\mathbf{b)}\; \left[\left(\frac{-1}{5}\right):\frac{1}{10}\right]- \frac{5}{7}\cdot \left(\frac{2}{3} – \frac{1}{5}\right)$$
$$= \left[ \frac{-1}{5} \cdot 10 \right] – \frac{5}{7} \cdot \left(\frac{10}{15} – \frac{3}{15}\right)$$
$$= (-2) – \frac{5}{7} \cdot \frac{7}{15}$$
$$= (-2) – \frac{1}{3}$$
$$= \frac{-6}{3} – \frac{1}{3} = \frac{-7}{3}.$$
$$\mathbf{c)}\; (-0,4)+ 2\frac{2}{5}\cdot \left[\left(\frac{-2}{3}\right)+\frac{1}{2}\right]^2$$
$$= \frac{-2}{5} + \frac{12}{5} \cdot \left[ \frac{-4}{6} + \frac{3}{6}\right]^2$$
$$= \frac{-2}{5} + \frac{12}{5} \cdot \left[\frac{-1}{6}\right]^2$$
$$= \frac{-2}{5} + \frac{12}{5} \cdot \frac{1}{36}$$
$$= \frac{-2}{5} + \frac{1}{15}$$
$$= \frac{-6}{15} + \frac{1}{15}$$
$$= \frac{-5}{15} = \frac{-1}{3}.$$
$$\mathbf{d)}\; \left\{ \left[ \left(\frac{1}{25}- 0,6\right)^2 : \frac{49}{125}\right]\cdot \frac{5}{6} \right\} – \left[\left(\frac{-1}{3}\right)+\frac{1}{2}\right]$$
$$= \left\{ \left[\left(\frac{1}{25} – \frac{3}{5}\right)^2 : \frac{49}{125}\right] \cdot \frac{5}{6} \right\} – \left[\frac{-2}{6} + \frac{3}{6}\right]$$
$$= \left\{ \left[\left(\frac{1}{25} – \frac{15}{25}\right)^2 : \frac{49}{125}\right] \cdot \frac{5}{6} \right\} – \frac{1}{6}$$
$$= \left\{ \left[ \left(\frac{-14}{25}\right)^2 : \frac{49}{125} \right] \cdot \frac{5}{6}\right\} – \frac{1}{6}$$
$$= \left\{ \left[ \frac{196}{625} : \frac{49}{125} \right] \cdot \frac{5}{6} \right\} – \frac{1}{6}$$
$$= \left\{\left[\frac{196}{625} \cdot \frac{125}{49}\right] \cdot \frac{5}{6}\right\} – \frac{1}{6}$$
$$= \left\{\frac{4}{5} \cdot \frac{5}{6}\right\} – \frac{1}{6}$$
$$= \frac{4}{6} – \frac{1}{6}$$
$$= \frac{3}{6} = \frac{1}{2}.$$
Bài tập 3 (Trang 25 / Toán 7 – tập 1 / Chân trời sáng tạo) Cho biểu thức:
$$A = \left(2 + \frac{1}{3} – \frac{2}{5}\right) – \left(7-\frac{3}{5}-\frac{4}{3}\right) – \left(\frac{1}{5}+\frac{5}{3}-4\right).$$
Hãy tính giá trị của $A$ theo hai cách:
a) Tính giá trị của từng biểu thức trong dấu ngoặc trước.
b) Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.
Giải
a) Tính giá trị của từng biểu thức trong dấu ngoặc trước.
$$A = \left(2 + \frac{1}{3} – \frac{2}{5}\right) – \left(7-\frac{3}{5}-\frac{4}{3}\right) – \left(\frac{1}{5}+\frac{5}{3}-4\right)$$
$$= \left(\frac{30}{15} + \frac{5}{15} – \frac{6}{15}\right) – \left(\frac{105}{15} – \frac{9}{15} – \frac{20}{15}\right) – \left(\frac{3}{15} + \frac{25}{15} – \frac{60}{15}\right)$$
$$= \frac{29}{15} – \frac{76}{15} – \frac{-32}{15}$$
$$= \frac{-15}{15} = -1.$$
b) Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.
$$A = \left(2 + \frac{1}{3} – \frac{2}{5}\right) – \left(7-\frac{3}{5}-\frac{4}{3}\right) – \left(\frac{1}{5}+\frac{5}{3}-4\right)$$
$$A = 2 + \frac{1}{3} – \frac{2}{5} – 7 + \frac{3}{5} + \frac{4}{3} – \frac{1}{5} – \frac{5}{3} + 4$$
$$= 2 – 7 + 4 + \frac{1}{3} + \frac{4}{3} – \frac{5}{3} – \frac{2}{5} + \frac{3}{5} – \frac{1}{5}$$
$$= (2 – 7 + 4) + \left(\frac{1}{3} + \frac{4}{3} – \frac{5}{3} \right) – \left(\frac{2}{5} – \frac{3}{5} + \frac{1}{5}\right)$$
$$= (-1) + \frac{0}{3} – \frac{0}{5}$$
$$= -1.$$
Bài tập 4 (Trang 25 / Toán 7 – tập 1 / Chân trời sáng tạo) Tìm $x,$ biết:
$$\mathbf{a)}\; x + \frac{3}{5} = \frac{2}{3};$$
$$\mathbf{b)}\; \frac{3}{7} – x = \frac{2}{5};$$
$$\mathbf{c)}\; \frac{4}{9} – \frac{2}{3} x = \frac{1}{3};$$
$$\mathbf{d)}\; \frac{3}{10} x – 1\frac{1}{2} = \left(\frac{-2}{7}\right):\frac{5}{14}.$$
Giải
$$\mathbf{a)}\; x + \frac{3}{5} = \frac{2}{3}$$
$$x = \frac{2}{3} – \frac{3}{5} = \frac{10}{15} – \frac{9}{15} = \frac{1}{15}.$$
$$\mathbf{b)}\; \frac{3}{7} – x = \frac{2}{5}$$
$$\frac{3}{7} – \frac{2}{5} = x$$
$$x = \frac{3}{7} – \frac{2}{5} = \frac{15}{35} – \frac{14}{35} = \frac{1}{35}.$$
$$\mathbf{c)}\; \frac{4}{9} – \frac{2}{3} x = \frac{1}{3}$$
$$\frac{4}{9} – \frac{1}{3} = \frac{2}{3} x$$
$$\frac{2}{3} x = \frac{4}{9} – \frac{1}{3} = \frac{4}{9} – \frac{3}{9} = \frac{1}{9}$$
Vậy:
$$\frac{2}{3}x = \frac{1}{9}$$
Suy ra:
$$x = \frac{1}{9} : \frac{2}{3} = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6}.$$
$$\mathbf{d)}\; \frac{3}{10} x – 1\frac{1}{2} = \left(\frac{-2}{7}\right):\frac{5}{14}$$
Ta có:
$$\left(\frac{-2}{7} \right) : \frac{5}{14} = \frac{-2}{7} \cdot \frac{14}{5} = \frac{-4}{5}.$$
$$1\frac{1}{2} = \frac{3}{2}.$$
Do đó, ta viết lại đề bài là:
$$\frac{3}{10}x – \frac{3}{2} = \frac{-4}{5}$$
$$\frac{3}{10}x = \frac{-4}{5} + \frac{3}{2} = \frac{-8}{10} + \frac{15}{10} = \frac{7}{10}$$
Vậy:
$$\frac{3}{10}x = \frac{7}{10}$$
Suy ra:
$$x = \frac{7}{10} : \frac{3}{10} = \frac{7}{10} \cdot \frac{10}{3} =\frac{7}{3}.$$
Bài tập 5 (Trang 25 / Toán 7 – tập 1 / Chân trời sáng tạo) Tìm $x,$ biết:
$$\mathbf{a)}\; \frac{2}{9} : x + \frac{5}{6} = 0,5;$$
$$\mathbf{b)}\; \frac{3}{4} – \left(x – \frac{2}{3}\right) = 1\frac{1}{3};$$
$$\mathbf{c)}\; 1\frac{1}{4} : \left(x – \frac{2}{3}\right) = 0,75;$$
$$\mathbf{d)}\; \left(-\frac{5}{6}x + \frac{5}{4}\right):\frac{3}{2} = \frac{4}{3}.$$
Giải
$$\mathbf{a)}\; \frac{2}{9} : x + \frac{5}{6} = 0,5$$
$$\frac{2}{9} : x = 0,5 – \frac{5}{6} = \frac{1}{2} – \frac{5}{6}$$
$$\;\;\; = \frac{3}{6} – \frac{5}{6} = \frac{-2}{6} = \frac{-1}{3}$$
Vậy:
$$\frac{2}{9} : x = \frac{-1}{3}$$
Do đó:
$$x = \frac{2}{9} : \frac{-1}{3} = \frac{2}{9} \cdot (-3) = \frac{-2}{3}.$$
$$\mathbf{b)}\; \frac{3}{4} – \left(x – \frac{2}{3}\right) = 1\frac{1}{3}$$
Thực hiện bỏ dấu ngoặc, ta được:
$$\frac{3}{4} – x + \frac{2}{3} = 1\frac{1}{3}$$
$$\frac{3}{4} + \frac{2}{3} – x = \frac{4}{3}$$
$$\frac{9}{12} + \frac{8}{12} – x = \frac{4}{3}$$
$$\frac{17}{12} – x = \frac{4}{3}$$
$$x = \frac{17}{12} – \frac{4}{3} = \frac{17}{12} – \frac{16}{12} = \frac{1}{12}.$$
$$\mathbf{c)}\; 1\frac{1}{4} : \left(x – \frac{2}{3}\right) = 0,75$$
$$x – \frac{2}{3} = 1\frac{1}{4} : 0,75$$
$$\;\;\; = \frac{5}{4} : \frac{3}{4} = \frac{5}{4} \cdot \frac{4}{3} = \frac{5}{3}$$
Vậy:
$$x – \frac{2}{3} = \frac{5}{3}$$
Suy ra:
$$x = \frac{5}{3} + \frac{2}{3} = \frac{7}{3}.$$
$$\mathbf{d)}\; \left(-\frac{5}{6}x + \frac{5}{4}\right):\frac{3}{2} = \frac{4}{3}$$
$$\frac{-5}{6}x + \frac{5}{4} = \frac{4}{3} \cdot \frac{3}{2} = 2$$
Vậy:
$$\frac{-5}{6}x + \frac{5}{4} = 2$$
Do đó:
$$\frac{-5}{6} x = 2 – \frac{5}{4} = \frac{3}{4}$$
Vậy:
$$\frac{-5}{6}x = \frac{3}{4}$$
Do đó:
$$x = \frac{3}{4} : \frac{-5}{6} = \frac{3}{4} \cdot \frac{-6}{5} = \frac{-9}{10}.$$
Bài tập 6 (Trang 25 / Toán 7 – tập 1 / Chân trời sáng tạo) Tính nhanh:
$$\mathbf{a)}\; \frac{13}{23}\cdot \frac{7}{11} + \frac{10}{23} \cdot \frac{7}{11};$$
$$\mathbf{b)}\; \frac{5}{9}\cdot \frac{23}{11} – \frac{1}{11} \frac{5}{9} + \frac{5}{9};$$
$$\mathbf{c)}\; \left[\left(-\frac{4}{9}\right)+ \frac{3}{5}\right] : \frac{13}{17} + \left(\frac{2}{5} – \frac{5}{9}\right) : \frac{13}{17};$$
$$\mathbf{d)}\; \frac{3}{16}: \left(\frac{3}{22}-\frac{3}{11}\right)+\frac{3}{16} : \left(\frac{1}{10} – \frac{2}{5}\right).$$
Giải
$$\mathbf{a)}\; \frac{13}{23}\cdot \frac{7}{11} + \frac{10}{23} \cdot \frac{7}{11}$$
$$= \frac{7}{11} \cdot \left(\frac{13}{23} + \frac{10}{23}\right)$$
$$= \frac{7}{11} \cdot \frac{23}{23}$$
$$= \frac{7}{11}.$$
$$\mathbf{b)}\; \frac{5}{9}\cdot \frac{23}{11} – \frac{1}{11}\cdot \frac{5}{9} + \frac{5}{9}$$
$$= \frac{5}{9} \cdot \left(\frac{23}{11} – \frac{1}{11} + 1\right)$$
$$= \frac{5}{9} \cdot \left(\frac{22}{11} + 1\right)$$
$$= \frac{5}{9} \cdot (2 + 1)$$
$$= \frac{5}{9} \cdot 3$$
$$= \frac{5}{3}.$$
$$\mathbf{c)}\; \left[\left(-\frac{4}{9}\right)+ \frac{3}{5}\right] : \frac{13}{17} + \left(\frac{2}{5} – \frac{5}{9}\right) : \frac{13}{17}$$
$$= \left[\left(-\frac{4}{9}\right) + \frac{3}{5} + \frac{2}{5} – \frac{5}{9}\right] : \frac{13}{17}$$
$$= \left[\frac{3}{5} + \frac{2}{5} – \frac{5}{9} – \frac{4}{9}\right] : \frac{13}{17}$$
$$= \left[\left(\frac{3}{5} + \frac{2}{5}\right) – \left(\frac{5}{9} + \frac{4}{9}\right)\right] : \frac{13}{17}$$
$$= \left[1 – 1\right] : \frac{13}{17}$$
$$= 0 : \frac{13}{17}$$
$$= 0.$$
$$\mathbf{d)}\; \frac{3}{16}: \left(\frac{3}{22}-\frac{3}{11}\right)+\frac{3}{16} : \left(\frac{1}{10} – \frac{2}{5}\right)$$
$$= \frac{3}{16} : \left(\frac{3}{22} – \frac{6}{22}\right) + \frac{3}{16} : \left(\frac{1}{10} – \frac{4}{10}\right)$$
$$= \frac{3}{16} : \frac{-3}{22} + \frac{3}{16} : \frac{-3}{10}$$
$$= \frac{3}{16} \cdot \frac{-22}{3} + \frac{3}{16} \cdot \frac{-10}{3}$$
$$= \frac{3}{16} \cdot \left(\frac{-22}{3} + \frac{-10}{3}\right)$$
$$= \frac{3}{16} \cdot \frac{-32}{3}$$
$$= -2.$$