[BT-T7-1.2#3] Bài tập THỨ TỰ THỰC HIỆN CÁC PHÉP TÍNH.

Sau đây là các bài tập TOÁN về THỨ TỰ THỰC HIỆN CÁC PHÉP TÍNH dành cho học sinh lớp 7. Trước khi làm bài tập, nên xem lại lý thuyết trong các bài liên quan:

Các dạng bài tập thường gặp:

Bài tập 1: Thực hiện phép tính:

a) $\dfrac{1}{2} + \dfrac{2}{3} – \dfrac{3}{4};$

b) $-\dfrac{20}{7} – \dfrac{3}{14} + \dfrac{32}{21};$

c) $1,2 – \dfrac{4}{5} – \dfrac{7}{2};$

d) $\dfrac{5}{-8} + \dfrac{7}{3} – 3,8.$

Bài tập 2: Tính:

a) $2 – \dfrac{2}{7}\cdot \dfrac{-1}{3};$

b) $14 : \dfrac{7}{3} + 5;$

c) $1,2\cdot \dfrac{7}{3} + \dfrac{4}{5}\cdot \dfrac{3}{2};$

d) $\dfrac{3}{4} : \dfrac{9}{2} – 6,3 : \dfrac{9}{5};$

e) $\dfrac{-2}{7}\cdot \dfrac{5}{-3} – \dfrac{5}{6} : \dfrac{-7}{4}.$

Bài tập 3: Tính:

a) $1 + 2^3 – \left(\dfrac{2}{3}\right)^2;$

b) $\dfrac{-7}{9} + \left(\dfrac{1}{3}\right)^3 – \dfrac{17}{27};$

c) $(0,4)^2 + \dfrac{3}{8}\cdot \dfrac{2}{15} – (0,5)^3;$

d) $3^2\cdot \dfrac{2}{25} – \left(\dfrac{6}{5}\right)^2.$

Bài tập 4: Thực hiện phép tính:

a) $(-1,3)^2 + \left(\dfrac{-4}{6}\right)^3 – 1;$

b) $\dfrac{(-4)^5}{(-2)^8} + \left(\dfrac{-1}{2}\right)^2\cdot \left(-\dfrac{5}{3}\right).$

Bài tập 5: Tính giá trị của mỗi biểu thức sau:

a) $\dfrac{2}{3} – 4\cdot \left(\dfrac{1}{2} + \dfrac{3}{4}\right);$

b) $\left(\dfrac{-1}{3} + \dfrac{5}{6}\right) \cdot 11 – 7;$

c) $\dfrac{-1}{24} – \left[ \dfrac{1}{4} – \left(\dfrac{1}{2} – \dfrac{7}{8}\right)\right];$

d) $\left(\dfrac{5}{7} – 1\dfrac{2}{5}\right) – \left[0,5 – \left( -\dfrac{2}{7}- 0,1\right)\right].$

Bài tập 6: Cho hai biểu thức: $A = -\dfrac{2}{3} – 4\cdot \left(\dfrac{1}{2}+ \dfrac{3}{4}\right)^2$ và $B = 2\cdot (-1)^6 + \left(\dfrac{3}{4}\right)^2 – \dfrac{3}{8}.$ Hãy tính $A – B.$

Bài tập 7: Cho $A = 12\cdot \left(\dfrac{2}{3} – \dfrac{5}{6}\right)^2$ và $B = 4\cdot \left(-\dfrac{1}{2}\right)^3 – 2\cdot \left(-\dfrac{1}{2}\right)^2 + 3\cdot \left(-\dfrac{1}{2}\right)+1.$ So sánh $A$ và $B.$

Bài tập 8: Cho $A = \left\{ 3\dfrac{17}{8} \cdot \left[ \dfrac{5}{2} – \left(\dfrac{1}{3} + \dfrac{2}{9}\right)\right]\right\} : \left[ \left(\dfrac{-1}{2}\right)+ 0,25\right]^2.$

a) $A$ là số dương hay số âm?

b) Tìm số hữu tỷ $x$ thỏa mãn $Ax = -\dfrac{10}{9}.$

Đáp án các bài tập:

Bài tập 1:

a) $\dfrac{1}{2} + \dfrac{2}{3} – \dfrac{3}{4}$ $= \dfrac{3}{6} + \dfrac{4}{6} – \dfrac{3}{4}$ $= \dfrac{7}{6} – \dfrac{3}{4}$ $= \dfrac{14}{12} – \dfrac{9}{12}$ $= \dfrac{5}{12}.$

b) $-\dfrac{20}{7} – \dfrac{3}{14} + \dfrac{32}{21}$ $= \dfrac{-40}{14} – \dfrac{3}{14} + \dfrac{32}{21}$ $= \dfrac{-43}{14} + \dfrac{32}{21}$ $= \dfrac{-129}{42} + \dfrac{64}{42}$ $= \dfrac{-65}{42}.$

c) $1,2 – \dfrac{4}{5} – \dfrac{7}{2}$ $= 1,2 – 0,8 – 3,5$ $= -3,1.$

d) $\dfrac{5}{-8} + \dfrac{7}{3} – 3,8$ $= (-0,625) + \dfrac{7}{3} – 3,8$ $= (-0,625) – 3,8 + \dfrac{7}{3}$ $= (-4,425) + \dfrac{7}{3}$ $= \dfrac{-4425}{1000}+\dfrac{7}{3}$ $= \dfrac{-177}{40}+\dfrac{7}{3}$ $= \dfrac{-531}{120} + \dfrac{280}{120}$ $= \dfrac{-251}{120}.$

Bài tập 2:

a) $2 – \dfrac{2}{7}\cdot \dfrac{-1}{3}$ $= 2 – \dfrac{-2}{21}$ $= 2 + \dfrac{2}{21}$ $= 2\dfrac{2}{21}.$

b) $14 : \dfrac{7}{3} + 5$ $= 14\cdot \dfrac{3}{7} + 5$ $= \dfrac{14\cdot 3}{7} + 5$ $= 6 + 5$ $= 11.$

c) $1,2\cdot \dfrac{7}{3} + \dfrac{4}{5}\cdot \dfrac{3}{2}$

Ta có: $1,2 = \dfrac{12}{10} = \dfrac{6}{5}.$

Do đó:

$1,2\cdot \dfrac{7}{3} + \dfrac{4}{5}\cdot \dfrac{3}{2}$ $= \dfrac{6}{5}\cdot \dfrac{7}{3} + \dfrac{4}{5}\cdot \dfrac{3}{2}$ $= \dfrac{6\cdot 7}{5\cdot 3} + \dfrac{4\cdot 3}{5\cdot 2}$ $= \dfrac{2\cdot 7}{5\cdot 1} + \dfrac{2\cdot 3}{5\cdot 1}$ $= \dfrac{14}{5} + \dfrac{6}{5}$ $= \dfrac{20}{5}$ $= 4.$

d) $\dfrac{3}{4} : \dfrac{9}{2} – 6,3 : \dfrac{9}{5}$ $= \dfrac{3}{4}\cdot \dfrac{2}{9} – \dfrac{63}{10}\cdot \dfrac{5}{9}$ $= \dfrac{3\cdot 2}{4\cdot 9} – \dfrac{63\cdot 5}{10\cdot 9}$ $= \dfrac{1\cdot 1}{2\cdot 3} – \dfrac{7\cdot 1}{2\cdot 1}$ $= \dfrac{1}{6} – \dfrac{7}{2}$ $= \dfrac{1}{6} – \dfrac{21}{6}$ $= \dfrac{-20}{6}$ $= \dfrac{-10}{3}.$

e) $\dfrac{-2}{7}\cdot \dfrac{5}{-3} – \dfrac{5}{6} : \dfrac{-7}{4}$ $= \dfrac{-10}{-21} – \dfrac{5}{6}\cdot \dfrac{-4}{7}$ $= \dfrac{10}{21} – \dfrac{5\cdot (-4)}{6\cdot 7}$ $= \dfrac{10}{21} – \dfrac{5\cdot (-2)}{3\cdot 7}$ $= \dfrac{10}{21} – \dfrac{-10}{21}$ $= \dfrac{10}{21} + \dfrac{10}{21}$ $= \dfrac{20}{21}.$

Bài tập 3:

a) $1 + 2^3 – \left(\dfrac{2}{3}\right)^2$ $= 1 + 8 – \dfrac{4}{9}$ $= 9 – \dfrac{4}{9}$ $= \dfrac{81}{9} – \dfrac{4}{9}$ $= \dfrac{77}{9}.$

b) $\dfrac{-7}{9} + \left(\dfrac{1}{3}\right)^3 – \dfrac{17}{27}$ $= \dfrac{-7}{9} + \dfrac{1}{27} – \dfrac{17}{27}$ $= \dfrac{-21}{27}+\dfrac{1}{27} – \dfrac{17}{27}$ $= \dfrac{-20}{27} – \dfrac{17}{27}$ $= \dfrac{-37}{27}.$

c) $(0,4)^2 + \dfrac{3}{8}\cdot \dfrac{2}{15} – (0,5)^3$ $= 0,16 + \dfrac{3\cdot 2}{8\cdot 15} – 0,125$ $= 0,16 + \dfrac{1\cdot 1}{4\cdot 5} – 0,125$ $= 0,16 + \dfrac{1}{20} – 0,125$ $= 0,16 + 0,05 – 0,125$ $= 0,085.$

d) $3^2\cdot \dfrac{2}{25} – \left(\dfrac{6}{5}\right)^2$ $= 9\cdot \dfrac{2}{25} – \dfrac{36}{25}$ $= \dfrac{18}{25} – \dfrac{36}{25}$ $= \dfrac{-18}{25}.$

Bài tập 4:

a) $(-1,3)^2 + \left(\dfrac{-4}{6}\right)^3 – 1$ $= \left(\dfrac{-13}{10}\right)^2 + \left(\dfrac{-2}{3}\right)^3 – 1$ $= \dfrac{(-13)^2}{10^2} + \dfrac{(-2)^3}{3^3} – 1$ $= \dfrac{169}{100} + \dfrac{-8}{27} – 1$ $= \dfrac{4563}{2700} + \dfrac{-800}{2700} – 1$ $= \dfrac{3763}{2700} – 1$ $= \dfrac{3763}{2700}- \dfrac{2700}{2700}$ $= \dfrac{1063}{2700}.$

b) $\dfrac{(-4)^5}{(-2)^8} + \left(\dfrac{-1}{2}\right)^2\cdot \left(-\dfrac{5}{3}\right)$ $= \dfrac{-4^5}{2^8} + \left(\dfrac{1}{2}\right)^2 \cdot \dfrac{-5}{3}$ $= \dfrac{-\left(2^2\right)^5}{2^8} + \dfrac{1}{4} \cdot \dfrac{-5}{3}$ $= \dfrac{-2^{10}}{2^8} + \dfrac{-5}{12}$ $= -2^2 – \dfrac{5}{12}$ $= -4 – \dfrac{5}{12}$ $= \dfrac{-48}{12} – \dfrac{5}{12}$ $= \dfrac{-53}{12}.$

Bài tập 5:

a) $\dfrac{2}{3} – 4\cdot \left(\dfrac{1}{2} + \dfrac{3}{4}\right)$ $= \dfrac{2}{3} – 4\cdot \left(\dfrac{2}{4} + \dfrac{3}{4}\right)$ $= \dfrac{2}{3} – 4\cdot \dfrac{5}{4}$ $= \dfrac{2}{3} – 5$ $= \dfrac{2}{3} – \dfrac{15}{3}$ $= \dfrac{-13}{3}.$

b) $\left(\dfrac{-1}{3} + \dfrac{5}{6}\right) \cdot 11 – 7$ $= \left(\dfrac{-2}{6} + \dfrac{5}{6}\right) \cdot 11 – 7$ $= \dfrac{3}{6} \cdot 11 – 7$ $= \dfrac{1}{2}\cdot 11 – 7$ $= \dfrac{11}{2} – 7$ $= \dfrac{11}{2} – \dfrac{14}{2}$ $= \dfrac{-3}{2}.$

c) $\dfrac{-1}{24} – \left[ \dfrac{1}{4} – \left(\dfrac{1}{2} – \dfrac{7}{8}\right)\right]$ $= \dfrac{-1}{24} – \left[\dfrac{1}{4} – \left(\dfrac{4}{8} – \dfrac{7}{8}\right)\right]$ $= \dfrac{-1}{24} – \left[\dfrac{1}{4} – \dfrac{-3}{8}\right]$ $= \dfrac{-1}{24} – \left[\dfrac{2}{8} + \dfrac{3}{8}\right]$ $= \dfrac{-1}{24} – \dfrac{5}{8}$ $= \dfrac{-1}{24} – \dfrac{15}{24}$ $= \dfrac{-16}{24}$ $= \dfrac{-2}{3}.$

d) $\left(\dfrac{5}{7} – 1\dfrac{2}{5}\right) – \left[0,5 – \left( -\dfrac{2}{7}- 0,1\right)\right]$ $= \left(\dfrac{5}{7} – \dfrac{7}{5}\right) – \left[\dfrac{1}{2} – \left(\dfrac{-2}{7} – \dfrac{1}{10}\right)\right]$ $= \left(\dfrac{25}{35} – \dfrac{49}{35}\right) – \left[\dfrac{1}{2} – \left(\dfrac{-20}{70} – \dfrac{7}{70}\right)\right]$ $= \dfrac{-24}{35} – \left[\dfrac{1}{2} – \dfrac{-27}{70}\right]$ $= \dfrac{-24}{35} – \left[\dfrac{35}{70} + \dfrac{27}{70}\right]$ $= \dfrac{-24}{35} – \dfrac{62}{70}$ $= \dfrac{-24}{35} – \dfrac{31}{35}$ $= \dfrac{-55}{35}$ $= \dfrac{-11}{7}.$

Bài tập 6:

Ta có:

+) $A = -\dfrac{2}{3} – 4\cdot \left(\dfrac{1}{2}+ \dfrac{3}{4}\right)^2$ $= -\dfrac{2}{3} – 4\cdot \left(\dfrac{2}{4} + \dfrac{3}{4}\right)^2$ $= \dfrac{-2}{3} – 4\cdot \left(\dfrac{5}{4}\right)^2$ $= \dfrac{-2}{3} – 4\cdot \dfrac{5^2}{4^2}$ $= \dfrac{-2}{3} – 4\cdot \dfrac{25}{16}$ $= \dfrac{-2}{3} – \dfrac{25}{4}$ $= \dfrac{-8}{12} – \dfrac{75}{12}$ $= \dfrac{-83}{12}.$

+) $B = 2\cdot (-1)^6 + \left(\dfrac{3}{4}\right)^2 – \dfrac{3}{8}$ $= 2\cdot 1 + \dfrac{3^2}{4^2} – \dfrac{3}{8}$ $= 2 + \dfrac{9}{16} – \dfrac{3}{8}$ $= \dfrac{32}{16} + \dfrac{9}{16} – \dfrac{6}{16}$ $= \dfrac{32 + 9 – 6}{16}$ $= \dfrac{35}{16}.$

Vậy:

$A – B = \dfrac{-83}{12} – \dfrac{35}{16}$ $= \dfrac{-332}{48} – \dfrac{105}{48}$ $= \dfrac{-437}{48}.$

Bài tập 7:

Ta có:

+) $A = 12\cdot \left(\dfrac{2}{3} – \dfrac{5}{6}\right)^2$ $= 12\cdot \left(\dfrac{4}{6} – \dfrac{5}{6}\right)^2$ $= 12\cdot \left(\dfrac{-1}{6}\right)^2$ $= 12\cdot \dfrac{(-1)^2}{6^2}$ $= 12\cdot \dfrac{1}{36}$ $= \dfrac{1}{3}.$

+) $B = 4\cdot \left(-\dfrac{1}{2}\right)^3 – 2\cdot \left(-\dfrac{1}{2}\right)^2 + 3\cdot \left(-\dfrac{1}{2}\right)+1$ $= 4\cdot \left(\dfrac{-1}{2}\right)^3 – 2\cdot \left(\dfrac{-1}{2}\right)^2 + 3\cdot \dfrac{-1}{2} + 1$ $= 4\cdot \dfrac{-1}{8} – 2\cdot \dfrac{1}{4} + \dfrac{-3}{2} + 1$ $= \dfrac{-1}{2} – \dfrac{1}{2} – \dfrac{3}{2} + 1$ $= -1 – \dfrac{3}{2} + 1$ $= -1 + 1 – \dfrac{3}{2}$ $= 0 – \dfrac{3}{2}$ $= \dfrac{-3}{2}.$

Ta thấy rằng $A = \dfrac{1}{3} > 0$ và $B = \dfrac{-3}{2} < 0$ nên $A > B.$

Bài tập 8: Cho $A = \left\{ 3\dfrac{17}{8} \cdot \left[ \dfrac{5}{2} – \left(\dfrac{1}{3} + \dfrac{2}{9}\right)\right]\right\} : \left[ \left(\dfrac{-1}{2}\right)+ 0,25\right]^2.$

a) Ta có:

$A = \left\{ 3\dfrac{17}{8} \cdot \left[ \dfrac{5}{2} – \left(\dfrac{1}{3} + \dfrac{2}{9}\right)\right]\right\} : \left[ \left(\dfrac{-1}{2}\right)+ 0,25\right]^2$ $= \left\{ \dfrac{41}{8}\cdot \left[ \dfrac{5}{2} – \left(\dfrac{3}{9} + \dfrac{2}{9}\right)\right]\right\} : \left[\dfrac{-2}{4} + \dfrac{1}{4}\right]^2$ $=\left\{ \dfrac{41}{8}\cdot \left[ \dfrac{5}{2} – \dfrac{5}{9}\right]\right\} : \left[\dfrac{-1}{4}\right]^2$ $= \left\{ \dfrac{41}{8}\cdot \left[ \dfrac{45}{18} – \dfrac{10}{18}\right]\right\} : \dfrac{1}{16}$ $= \left\{ \dfrac{41}{8}\cdot \dfrac{35}{18}\right\} : \dfrac{1}{16}$ $= \dfrac{41\cdot 35}{8\cdot 18}\cdot 16$ $= \dfrac{41\cdot 35\cdot 16}{8\cdot 18}$ $= \dfrac{41\cdot 35}{9}$ $= \dfrac{1435}{9}.$

Vậy $A$ là số dương.

b) $Ax = -\dfrac{10}{9}$ tức là $\dfrac{1435}{9}\cdot x = -\dfrac{10}{9}$

Do đó:

$x = -\dfrac{10}{9} : \dfrac{1435}{9}$ $= \dfrac{-10}{9} \cdot \dfrac{9}{1435}$ $= \dfrac{-10}{1435}.$