Bài 1.8 trang 13 Toán lớp 7: Tính giá trị của các biểu thức sau:

$\begin{array}{l}a)(8+2\frac{1}{3}-\frac{3}{5})-(5+0,4)-(3\frac{1}{3}-2)\\ b)(7-\frac{1}{2}-\frac{3}{4}):(5-\frac{1}{4}-\frac{5}{8})\end{array}$

Phương pháp giải:

a) Cách 1:

$\begin{array}{l}(8+2\frac{1}{3}-\frac{3}{5})-(5+0,4)-(3\frac{1}{3}-2)\\ =(8+\frac{7}{3}-\frac{3}{5})-(5+\frac{4}{10})-(\frac{10}{3}-2)\\ =8+\frac{7}{3}-\frac{3}{5}-5-\frac{2}{5}-\frac{10}{3}+2\\ =(8-5+2)+(\frac{7}{3}-\frac{10}{3})-(\frac{3}{5}+\frac{2}{5})\\ =5+\frac{-3}{3}-\frac{5}{5}\\ =5+(-1)-1\\ =3\end{array}$

Cách 2:

$\begin{array}{l}(8+2\frac{1}{3}-\frac{3}{5})-(5+0,4)-(3\frac{1}{3}-2)\\ =(8+\frac{7}{3}-\frac{3}{5})-(5+\frac{4}{10})-(\frac{10}{3}-2)\\ =(\frac{120}{15}+\frac{35}{15}-\frac{9}{15})-(\frac{25}{5}+\frac{2}{5})-(\frac{10}{3}-\frac{6}{3})\\ =\frac{146}{15}-\frac{27}{5}-\frac{4}{3}\\ =\frac{146}{15}-\frac{81}{15}-\frac{20}{15}\\ =\frac{45}{15}\\ =3\end{array}$

b)

$\begin{array}{l}(7-\frac{1}{2}-\frac{3}{4}):(5-\frac{1}{4}-\frac{5}{8})\\ =(\frac{28}{4}-\frac{2}{4}-\frac{3}{4}):(\frac{40}{8}-\frac{2}{8}-\frac{5}{8})\\ =\frac{23}{4}:\frac{33}{8}\\ =\frac{23}{4}.\frac{8}{33}\\ =\frac{46}{33}\end{array}$

Bài tập vận dụng:

Bài 1. Tính giá trị của các biểu thức sau:

a) $\left(7+3\frac{1}{4}-\frac{3}{5}\right)+\left(0,4-5\right)-\left(4\frac{1}{4}-1\right)$;

b) $\frac{2}{3}-\left[\left(-\frac{7}{4}\right)-\left(\frac{1}{2}+\frac{3}{8}\right)\right]$;

c) $\left(9-\frac{1}{2}-\frac{3}{4}\right):\left(7-\frac{1}{4}-\frac{5}{8}\right)$;

d) $3-\frac{1-\frac{1}{7}}{1+\frac{1}{7}}$.

Hướng dẫn giải

a) $\left(7+3\frac{1}{4}-\frac{3}{5}\right)+\left(0,4-5\right)-\left(4\frac{1}{4}-1\right)$

$=7+3\frac{1}{4}-\frac{3}{5}+0,4-5-4\frac{1}{4}+1$

$=\left(7-5+1\right)+\left(3\frac{1}{4}-4\frac{1}{4}\right)+\left(-\frac{3}{5}+0,4\right)$

$=3+\left(-1\right)+\left(-0,6+0,4\right)$

$=2+\left(-0,2\right)=1,8$

b) $\frac{2}{3}-\left[\left(-\frac{7}{4}\right)-\left(\frac{1}{2}+\frac{3}{8}\right)\right]$

$=\frac{2}{3}-\left[\left(-\frac{7}{4}\right)-\left(\frac{4}{8}+\frac{3}{8}\right)\right]$

$=\frac{2}{3}-\left[\left(-\frac{14}{8}\right)-\frac{7}{8}\right]$

$=\frac{2}{3}-\left(-\frac{21}{8}\right)$

$=\frac{2}{3}+\frac{21}{8}$

$=\frac{16}{24}+\frac{63}{24}=\frac{79}{24}$

c) $\left(9-\frac{1}{2}-\frac{3}{4}\right):\left(7-\frac{1}{4}-\frac{5}{8}\right)$

$=\left(\frac{36}{4}-\frac{2}{4}-\frac{3}{4}\right):\left(\frac{56}{8}-\frac{2}{8}-\frac{5}{8}\right)$

$=\frac{31}{4}:\frac{49}{8}=\frac{31}{4}\cdot \frac{8}{49}=\frac{62}{49}$

d) $\left(1+\frac{1}{7}\right)=3-\frac{6}{7}:\frac{8}{7}=3-\frac{6}{7}\cdot \frac{7}{8}=3-\frac{3}{4}=\frac{12}{4}-\frac{3}{4}=\frac{9}{4}$.

Bài 2. Tính:

a) $\frac{-8}{18}+\frac{15}{27}$;

b) $3,5-\left(-\frac{2}{7}\right)$;

c) $-0,25\cdot \left(-0,4\right)$;

d) $-6:3\frac{1}{5}$.

Hướng dẫn giải

a) $\frac{-8}{18}+\frac{15}{27}=\frac{-4}{9}+\frac{5}{9}=\frac{1}{9}$;

b) $3,5-\left(-\frac{2}{7}\right)=\frac{7}{2}+\frac{2}{7}=\frac{49}{14}+\frac{4}{14}=\frac{53}{14}$;

c) $-0,25\cdot \left(-0,4\right)=0,25\cdot 0,4=0,1$

d) $-6:3\frac{1}{5}=-6:\frac{16}{5}=-6\cdot \frac{5}{16}=\frac{-15}{8}$.

Bài 3. Giá trị của biểu thức $\frac{3}{5}-\left(\frac{1}{2}-\frac{2}{5}\right)$ là

A. $\frac{-3}{10}$;

B. $\frac{1}{2}$;

C. $\frac{3}{2}$;

D. $\frac{7}{10}$.

Hướng dẫn giải

Đáp án đúng là: B

$\frac{3}{5}-\left(\frac{1}{2}-\frac{2}{5}\right)$ = $\frac{3}{5}-\frac{1}{2}+\frac{2}{5}$ = $\left(\frac{3}{5}+\frac{2}{5}\right)-\frac{1}{2}$ = $1-\frac{1}{2}$ = $\frac{1}{2}$.