Bài tập TOÁN 7 (CT mới) – Chuyên đề THỨ TỰ THỰC HIỆN CÁC PHÉP TÍNH. QUY TẮC DẤU NGOẶC VÀ QUY TẮC CHUYỂN VẾ.

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

$=\dfrac{6}{12}+\dfrac{8}{12}-\dfrac{9}{12}$

$=\dfrac{6+8-9}{12}=\dfrac{5}{12}.$

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) $1,2-\dfrac{4}{5}-\dfrac{7}{2}$

$=\dfrac{6}{5}-\dfrac{4}{5}-\dfrac{7}{2}$

$=\dfrac{2}{5}-\dfrac{7}{2}$

$=\dfrac{4}{10}-\dfrac{35}{10}$

$=\dfrac{-31}{10}=-3,1.$

a) $2-\dfrac{2}{7}\cdot\dfrac{-1}{3}$ $=2-\dfrac{-2}{21}$ $=\dfrac{42}{21}-\dfrac{-2}{21}$ $=\dfrac{42}{21}+\dfrac{2}{21}$ $=\dfrac{44}{21}.$

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

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

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

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

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

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

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

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

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) $\left(\dfrac{-1}{3}+\dfrac{5}{6}\right)\cdot 11-7$ $=\left(\dfrac{-2}{6}+\dfrac{5}{6}\right)\cdot 11-7$ $=\dfrac{3}{6}\cdot 11-7$ $=\dfrac{33}{6}-7$ $=\dfrac{33}{6}-\dfrac{42}{6}$ $=\dfrac{-9}{6}=\dfrac{-3}{2}.$

c) $\dfrac{15}{16}:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^2$ $=\dfrac{15}{16}:\left(\dfrac{3}{6}-\dfrac{4}{6}\right)^2$ $=\dfrac{15}{16}:\left(\dfrac{-1}{6}\right)^2$ $=\dfrac{15}{16}:\dfrac{1}{36}$ $=\dfrac{15}{16}\cdot 36$ $=\dfrac{15\cdot 9}{4}=\dfrac{135}{4}.$

d) $12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}$ $=12\cdot\dfrac{4}{9}+\dfrac{4}{3}$ $=\dfrac{4\cdot 4}{3}+\dfrac{4}{3}$ $=\dfrac{16}{3}+\dfrac{4}{3}$ $=\dfrac{20}{3}.$

e) $\left(-\dfrac{1}{7}\right)^0-\left(2\dfrac{4}{9}\right)\cdot\left(\dfrac{3}{2}\right)^2$ $=1-\dfrac{22}{9}\cdot\dfrac{9}{4}$ $=1-\dfrac{11}{2}$ $=\dfrac{2}{2}-\dfrac{11}{2}$ $=\dfrac{-9}{2}.$

a) $\dfrac{-4}{27}-\dfrac{7}{12}+\dfrac{-23}{27}+\dfrac{17}{12}$

$=\dfrac{-4}{27}+\dfrac{-23}{27}-\dfrac{7}{12}+\dfrac{17}{12}$

$=\left(\dfrac{-4}{27}+\dfrac{-23}{27}\right)-\left(\dfrac{7}{12}-\dfrac{17}{12}\right)$

$=\dfrac{-27}{27}-\dfrac{-10}{12}$

$=-1+\dfrac{5}{6}$

$=\dfrac{-6}{6}+\dfrac{5}{6}=\dfrac{-1}{6}.$

b) $\dfrac{5}{23}+\dfrac{7}{17}-0,25-\dfrac{5}{23}+\dfrac{10}{17}$

$=\dfrac{5}{23}-\dfrac{5}{23}+\dfrac{7}{17}+\dfrac{10}{17}-0,25$

$=\left(\dfrac{5}{23}-\dfrac{5}{23}\right)+\left(\dfrac{7}{17}+\dfrac{10}{17}\right)-0,25$

$=0+\dfrac{17}{17}-0,25$

$=0+1-0,25=0,75.$

b) $\dfrac{-2}{5}\cdot\dfrac{9}{7}+\dfrac{9}{7}\cdot\dfrac{-3}{10}$

$=\dfrac{9}{7}\cdot\left(\dfrac{-2}{5}+\dfrac{-3}{10}\right)$

$=\dfrac{9}{7}\cdot\left(\dfrac{-4}{10}+\dfrac{-3}{10}\right)$

$=\dfrac{9}{7}\cdot\dfrac{-7}{10}=\dfrac{-9}{10}.$

c) $\dfrac{5}{9}\cdot\dfrac{11}{8}-\dfrac{5}{9}\cdot\dfrac{-19}{8}$

$=\dfrac{5}{9}\cdot\left(\dfrac{11}{8}-\dfrac{-19}{8}\right)$

$=\dfrac{5}{9}\cdot\dfrac{30}{18}$

$=\dfrac{5}{9}\cdot\dfrac{5}{3}=\dfrac{25}{27}.$

d) $\left(-\dfrac{25}{13}\right)-\left(\dfrac{25}{17}-\dfrac{12}{13}+\dfrac{-8}{17}\right)$

$=-\dfrac{25}{13}-\dfrac{25}{17}+\dfrac{12}{13}-\dfrac{-8}{17}$

$=\dfrac{-25}{13}-\dfrac{25}{17}+\dfrac{12}{13}+\dfrac{8}{17}$

$=\dfrac{-25}{13}+\dfrac{12}{13}-\dfrac{25}{17}+\dfrac{8}{17}$

$=\left(\dfrac{-25}{13}+\dfrac{12}{13}\right)-\left(\dfrac{25}{17}-\dfrac{8}{17}\right)$

$=\dfrac{-13}{13}-\dfrac{17}{17}$

$=-1-1=-2.$

Giá trước khi giảm của $4$ cái bánh pizza là: $4\cdot 19=76$ (USD).

Số tiền được giảm khi mua $4$ cái bánh pizza là: $4\cdot 1,5=6$ (USD).

Suy ra tổng số tiền chị Trang dùng để mua bánh là: $76-6=70$ (USD).

Giá của $2$ cái bánh cỡ to, $3$ cái bánh cỡ trung, $3$ cái bánh cỡ nhỏ là: $2\cdot 11,5+3\cdot 9,5+3\cdot 6,25=70,25$ (đô la).

Số tiền người bán hàng phải trả lại cho Linda là: $100-70,25=29,75$ (đô la).

a) $5x-9=5+3x$

$5x-3x=5+9$

$2x=14$

$x=14:2=7.$

Vậy $x=7.$

b) $\dfrac{2}{5}-x=\dfrac{1}{2}x-2$

$\dfrac{2}{5}+2=\dfrac{1}{2}x+x$

$\dfrac{12}{5}=\dfrac{3}{2}x$

$x=\dfrac{12}{5}:\dfrac{3}{2}=\dfrac{12}{5}\cdot\dfrac{2}{3}=\dfrac{8}{5}.$

Vậy $x=\dfrac{8}{5}.$

a) $1-\left(-x+\dfrac{9}{5}\right)=\dfrac{5}{6}+\left(-\dfrac{7}{12}\right)$

Ta có:

+) $1-\left(-x+\dfrac{9}{5}\right)$ $=1+x-\dfrac{9}{5}$ $=1-\dfrac{9}{5}+x$ $=\dfrac{5}{5}-\dfrac{9}{5}+x$ $=-\dfrac{4}{5}+x.$

+) $\dfrac{5}{6}+\left(-\dfrac{7}{12}\right)$ $=\dfrac{5}{6}-\dfrac{7}{12}$ $=\dfrac{10}{12}-\dfrac{7}{12}$ $=\dfrac{3}{12}$ $=\dfrac{1}{4}.$

Vậy: $-\dfrac{4}{5}+x=\dfrac{1}{4}.$

Do đó: $x=\dfrac{1}{4}+\dfrac{4}{5}=\dfrac{-11}{20}.$

b) $1-\left(\dfrac{8}{7}-x\right)=\dfrac{4}{5}+\dfrac{-1}{10}.$

Ta có:

+) $1-\left(\dfrac{8}{7}-x\right)$ $=1-\dfrac{8}{7}+x$ $=-\dfrac{1}{7}+x.$

+) $\dfrac{4}{5}+\dfrac{-1}{10}$ $=\dfrac{8}{10}-\dfrac{1}{10}$ $=\dfrac{7}{10}.$

Vậy $-\dfrac{1}{7}+x=\dfrac{7}{10}.$

Do đó: $x=\dfrac{7}{10}+\dfrac{1}{7}=\dfrac{59}{70}.$

c) $\left(-0,75\cdot x+\dfrac{5}{2}\right)\cdot\dfrac{4}{7}=\dfrac{-1}{3}$

Ta có: $\left(-0,75\cdot x+\dfrac{5}{2}\right)\cdot\dfrac{4}{7}$ $=\left(\dfrac{-3}{4}x+\dfrac{5}{2}\right)\cdot\dfrac{4}{7}$ $=\dfrac{-3}{4}x\cdot\dfrac{4}{7}+\dfrac{5}{2}\cdot\dfrac{4}{7}$ $=-\dfrac{3}{7}x+\dfrac{10}{7}.$

Vậy $-\dfrac{3}{7}x+\dfrac{10}{7}=\dfrac{-1}{3}.$

Do đó:

$\dfrac{1}{3}+\dfrac{10}{7}=\dfrac{3}{7}x$

$\dfrac{7}{21}+\dfrac{30}{21}=\dfrac{3}{7}x$

$\dfrac{37}{21}=\dfrac{3}{7}x$

$x=\dfrac{37}{21}:\dfrac{3}{7}$

$x=\dfrac{37}{21}\cdot\dfrac{7}{3}=\dfrac{37}{9}.$

Vậy $x=\dfrac{37}{9}.$

d) $\dfrac{2}{5}-3:x=2\cdot\left(\dfrac{1}{5}\right)^2$

Ta có: $2\cdot\left(\dfrac{1}{5}\right)^2$ $=2\cdot\dfrac{1}{25}$ $=\dfrac{2}{25}.$

Vậy $\dfrac{2}{5}-3:x=\dfrac{2}{25}.$

Do đó: $3:x=\dfrac{2}{5}-\dfrac{2}{25}=\dfrac{10}{25}-\dfrac{2}{25}=\dfrac{8}{25}.$

Dẫn đến $x=3:\dfrac{8}{25}=3\cdot\dfrac{25}{8}=\dfrac{75}{8}.$

e) $2^3+0,5\cdot x=1,5$

$8+0,5\cdot x=1,5$

$0,5\cdot x=1,5-8$

$0,5\cdot x=-6,5$

$x=-6,5:0,5$

$x=13.$

a) $\dfrac{5}{7}\cdot\left(\dfrac{-3}{11}\right)+\dfrac{5}{7}\cdot\left(\dfrac{-8}{11}\right)+2\dfrac{5}{7}$

$=\dfrac{5}{7}\cdot\left(\dfrac{-3}{11}+\dfrac{-8}{11}\right)+2\dfrac{5}{7}$

$=\dfrac{5}{7}\cdot\dfrac{-11}{11}+\left(2+\dfrac{5}{7}\right)$

$=\dfrac{5}{7}\cdot(-1)+2+\dfrac{5}{7}$

$=\dfrac{-5}{7}+2+\dfrac{5}{7}$

$=\dfrac{-5}{7}+\dfrac{5}{7}+2$

$=0+2=2.$

b) $\dfrac{-3}{17}:\dfrac{5}{4}+\dfrac{-3}{17}:3$

$=\dfrac{-3}{17}\cdot\dfrac{4}{5}+\dfrac{-3}{17}\cdot\dfrac{1}{3}$

$=\dfrac{-3}{17}\cdot\left(\dfrac{4}{5}+\dfrac{1}{3}\right)$

$=\dfrac{-3}{17}\cdot\left(\dfrac{12}{15}+\dfrac{5}{15}\right)$

$=\dfrac{-3}{17}\cdot\dfrac{17}{15}$

$=\dfrac{-1}{5}.$

c) $\dfrac{3}{4}-\left[\left(-\dfrac{5}{3}\right)-\left(\dfrac{1}{12}-\dfrac{5}{3}\right)\right]$

$=\dfrac{3}{4}+\dfrac{5}{3}+\dfrac{1}{12}+\dfrac{5}{3}$

$=\dfrac{3}{4}+\dfrac{1}{12}+\dfrac{5}{3}+\dfrac{5}{3}$

$=\left(\dfrac{3}{4}+\dfrac{1}{12}\right)+\left(\dfrac{5}{3}+\dfrac{5}{3}\right)$

$=\left(\dfrac{9}{12}+\dfrac{1}{12}\right)+\dfrac{10}{3}$

$=\dfrac{10}{12}+\dfrac{10}{3}$

$=\dfrac{5}{6}+\dfrac{20}{6}$

$=\dfrac{25}{6}.$

d) $\dfrac{3}{4}\cdot\left(26\dfrac{2}{9}\right)-\left(38\dfrac{2}{9}\right)\cdot\dfrac{3}{4}$

$=\dfrac{3}{4}\cdot\left(26\dfrac{2}{9}-38\dfrac{2}{9}\right)$

$=\dfrac{3}{4}\cdot\left[\left(26+\dfrac{2}{9}\right)-\left(38+\dfrac{2}{9}\right)\right]$

$=\dfrac{3}{4}\cdot\left[26+\dfrac{2}{9}-38-\dfrac{2}{9}\right]$

$=\dfrac{3}{4}\cdot\left[26-38\right]$

$=\dfrac{3}{4}\cdot(-12)=-9.$

e) $\left(35\dfrac{1}{6}\right):\left(-\dfrac{4}{5}\right)-45\dfrac{1}{6}:\dfrac{-4}{5}$

$=\left(35\dfrac{1}{6}-45\dfrac{1}{6}\right):\dfrac{-4}{5}$

$=\left[\left(35+\dfrac{1}{6}\right)-\left(45+\dfrac{1}{6}\right)\right]\cdot\dfrac{-5}{4}$

$=\left[35+\dfrac{1}{6}-45-\dfrac{1}{6}\right]\cdot\dfrac{-5}{4}$

$=[35-45]\cdot\dfrac{-5}{4}$

$=(-10)\cdot\dfrac{-5}{4}=\dfrac{25}{2}.$

f) $\left(-\dfrac{2}{3}\right)^3:\left(\dfrac{-2}{3}\right)^2+\dfrac{2^{40}\cdot 3^{29}}{8^{13}\cdot 9^{15}}$

Ta có:

+) $\left(-\dfrac{2}{3}\right)^3:\left(\dfrac{-2}{3}\right)^2$ $=\left(\dfrac{-2}{3}\right)^3:\left(\dfrac{-2}{3}\right)^2$ $=\left(\dfrac{-2}{3}\right)^{3-2}$ $=\left(\dfrac{-2}{3}\right)^{1}$ $=\dfrac{-2}{3}.$

+) $\dfrac{2^{40}\cdot 3^{29}}{8^{13}\cdot 9^{15}}$ $=\dfrac{2^{40}\cdot 3^{29}}{\left(2^3\right)^{13}\cdot\left(3^2\right)^{15}}$ $=\dfrac{2^{40}\cdot 3^{29}}{2^{3\cdot 13}\cdot 3^{2\cdot 15}}$ $=\dfrac{2^{40}\cdot 3^{29}}{2^{39}\cdot 3^{30}}$ $=\dfrac{2^{1}}{3^{1}}$ $=\dfrac{2}{3}.$

Vậy:

$\left(-\dfrac{2}{3}\right)^3:\left(\dfrac{-2}{3}\right)^2+\dfrac{2^{40}\cdot 3^{29}}{8^{13}\cdot 9^{15}}$ $=\dfrac{-2}{3}+\dfrac{2}{3}=0.$

Cách 1:

Món hàng thứ nhất được giảm $30\%$ nên số tiền phải trả chỉ bằng $70\%$ giá tiền. Vậy số tiền để mua món hàng thứ nhất là: $125\;000\cdot\dfrac{70}{100}=87\;500$ (đồng).

Món hàng thứ hai được giảm $15\%$ nên số tiền phải trả chỉ bằng $85\%$ giá tiền. Vậy số tiền để mua món hàng thứ hai là: $300\;000\cdot\dfrac{85}{100}=255\;000$ (đồng).

Tổng số tiền phải trả (cho cả ba món hàng) là $692\;000$ đồng nên số tiền để mua món hàng thứ ba là: $692\;000-87\;500-255\;000=349\;500$ (đồng).

Vì món hàng thứ ba đã được giảm giá $40\%$ nên số tiền phải trả chỉ bằng $60\%$ giá tiền. Vậy giá tiền của món hàng thứ ba là: $349\;500\cdot\dfrac{100}{60}=582\;500$ (đồng).

Cách 2:

Gọi $x$ (đồng) là giá tiền (lúc chưa giảm giá) của món hàng thứ ba.

Số tiền để mua món hàng thứ nhất là $125\;000\cdot\dfrac{70}{100}=87\;500$ (đồng).

Số tiền để mua món hàng thứ hai là $300\;000\cdot\dfrac{85}{100}=255\;000$ (đồng).

Số tiền để mua món hàng thứ ba là $x\cdot\dfrac{60}{100}=x\cdot\dfrac{3}{5}$ (đồng).

Theo đề ta có: $87\;500+255\;000+x\cdot\dfrac{3}{5}=692\;000$

Do đó: $x\cdot\dfrac{3}{5}=692\;000-87\;500-255\;000=349\;500$

Suy ra: $x=349\;500:\dfrac{3}{5}=349\;500\cdot\dfrac{5}{3}=582\;500$

Vậy món hàng thứ ba có giá $582\;500$ đồng.

Tổng số tiền được giảm giá là $4\cdot 1,5=6$ (USD).

Số tiền chị Trang dùng để mua bánh là $41-6=35$ (USD).

Ta có: $\dfrac{7}{2}-\left[\dfrac{3}{2}-\left(x+\dfrac{7}{2}\right)\right]$ $=\dfrac{7}{2}-\dfrac{3}{2}+\left(x+\dfrac{7}{2}\right)$ $=\dfrac{7}{2}-\dfrac{3}{2}+x+\dfrac{7}{2}$ $=\left(\dfrac{7}{2}-\dfrac{3}{2}+\dfrac{7}{2}\right)+x$ $=\dfrac{11}{2}+x$

Vậy $\dfrac{11}{2}+x=\dfrac{-9}{11}$

Do đó $x=\dfrac{-9}{11}-\dfrac{11}{2}=\dfrac{(-9)\cdot 2-11\cdot 11}{11\cdot 2}=\dfrac{-139}{22}.$

Ta có: $\dfrac{x+2}{338}+\dfrac{x+3}{337}+\dfrac{x+4}{336}+\dfrac{x+5}{335}+\dfrac{x+360}{5}$ $=\left(\dfrac{x+2}{338}+1\right)+\left(\dfrac{x+3}{337}+1\right)+\left(\dfrac{x+4}{336}+1\right)+\left(\dfrac{x+5}{335}+1\right)+\left(\dfrac{x+360}{5}-4\right)$ $=\dfrac{x+2+338}{338}+\dfrac{x+3+337}{337}+\dfrac{x+4+336}{336}+\dfrac{x+5+335}{335}+\dfrac{x+360-4\cdot 5}{5}$ $=\dfrac{x+340}{338}+\dfrac{x+340}{337}+\dfrac{x+340}{336}+\dfrac{x+340}{335}+\dfrac{x+340}{5}$ $=(x+340)\cdot\left(\dfrac{1}{338}+\dfrac{1}{337}+\dfrac{1}{336}+\dfrac{1}{335}+\dfrac{1}{5}\right).$

Vậy $(x+340)\cdot\left(\dfrac{1}{338}+\dfrac{1}{337}+\dfrac{1}{336}+\dfrac{1}{335}+\dfrac{1}{5}\right)=0$

Suy ra $x+340=0,$ hay $x=-340.$