Bài 2 trang 25 Toán lớp 7: Tính:

a) $\left(\frac{3}{4}:1\frac{1}{2}\right)-\left(\frac{5}{6}:\frac{1}{3}\right)$                  b) $\left[\left(\frac{-1}{5}\right):\frac{1}{10}\right]-\frac{5}{7}.\left(\frac{2}{3}-\frac{1}{5}\right)$

c) $\left(-0,4\right)+2\frac{2}{5}.{\left[\left(\frac{-2}{3}\right)+\frac{1}{2}\right]}^2$      d)$\left\{\left[{\left(\frac{1}{25}-0,6\right)}^2:\frac{49}{125}\right].\frac{5}{6}\right\}-\left[\left(\frac{-1}{3}\right)+\frac{1}{2}\right]$

Phương pháp giải:

Lời giải:

a)

$\begin{array}{l}\left(\frac{3}{4}:1\frac{1}{2}\right)-\left(\frac{5}{6}:\frac{1}{3}\right)\\ =\left(\frac{3}{4}:\frac{3}{2}\right)-\left(\frac{5}{6}.3\right)\\ =\left(\frac{3}{4}.\frac{2}{3}\right)-\frac{5}{2}\\ =\frac{1}{2}-\frac{5}{2}\\ =-2.\end{array}$                         

b)

$\begin{array}{l}\left[\left(\frac{-1}{5}\right):\frac{1}{10}\right]-\frac{5}{7}.\left(\frac{2}{3}-\frac{1}{5}\right)\\ =\left(\frac{-1}{5}\right).10-\frac{5}{7}.\left(\frac{10}{15}-\frac{3}{15}\right)\\ =-2-\frac{5}{7}.\frac{7}{15}\\ =-2-\frac{1}{3}\\ =\frac{-6}{3}-\frac{1}{3}\\ =\frac{-7}{3}\end{array}$

c)

$\begin{array}{l}\left(-0,4\right)+2\frac{2}{5}.{\left[\left(\frac{-2}{3}\right)+\frac{1}{2}\right]}^2\\ =\left(-\frac{2}{5}\right)+\frac{12}{5}.{\left[\left(\frac{-4}{6}\right)+\frac{3}{6}\right]}^2\\ =\left(-\frac{2}{5}\right)+\frac{12}{5}.{\left(\frac{-1}{6}\right)}^2\\ =\left(-\frac{2}{5}\right)+\frac{12}{5}.\frac{1}{36}\\ =\left(-\frac{2}{5}\right)+\frac{1}{15}\\ =\left(-\frac{6}{15}\right)+\frac{1}{15}\\ =\frac{-5}{15}=\frac{-1}{3}\end{array}$             

d)

$\begin{array}{l}\left\{\left[{\left(\frac{1}{25}-0,6\right)}^2:\frac{49}{125}\right].\frac{5}{6}\right\}-\left[\left(\frac{-1}{3}\right)+\frac{1}{2}\right]\\ =\left\{\left[{\left(\frac{1}{25}-\frac{3}{5}\right)}^2.\frac{125}{49}\right].\frac{5}{6}\right\}-\left[\left(\frac{-2}{6}\right)+\frac{3}{6}\right]\\ =\left\{\left[{\left(\frac{-14}{25}\right)}^2.\frac{125}{49}\right].\frac{5}{6}\right\}-\frac{1}{6}\\ =\left\{\left[{\left(\frac{-14}{25}\right)}^2.\frac{125}{49}\right].\frac{5}{6}\right\}-\frac{1}{6}\\ =\left\{\frac{196}{{25}^2}.\frac{25.5}{49}.\frac{5}{6}\right\}-\frac{1}{6}\\ =\left(\frac{\mathrm{4.49.25.5.5}}{{25}^2.49.6}\right)-\frac{1}{6}\\ =\frac{4}{6}-\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\end{array}$

Bài tập vận dụng:

Câu 1. Kết quả của phép tính $\frac{1}{2023}.\frac{-7}{9}+\frac{2022}{2023}.\frac{-7}{9}+\frac{7}{9}$ bằng:

A. 10;

B. 20;

C. 0;

D. 15.

Hướng dẫn giải:

Đáp án đúng là: C

Ta có $\frac{1}{2023}.\frac{-7}{9}+\frac{2022}{2023}.\frac{-7}{9}+\frac{7}{9}$

 $=\frac{1}{2023}.\frac{-7}{9}+\frac{2022}{2023}.\frac{-7}{9}+\frac{-7}{9}.\left(-1\right)$

 $=\frac{-7}{9}.\left(\frac{1}{2023}+\frac{2022}{2023}-1\right)$

$=\frac{-7}{9}.\left(\frac{2023}{2023}-1\right)=\frac{-7}{9}.\left(1-1\right)=\frac{-7}{9}.0=0$.

Vậy ta chọn phương án C.

Câu 2. Số hữu tỉ x thỏa mãn $\frac{7}{4}-\left(\frac{x}{60}+\frac{5}{3}\right)=\frac{-12}{5}$ là:

A. x = 149;

B. x = ‒149;

C. x = $\frac{149}{6}$ ;

D. $x=-\frac{149}{6}$.

Hướng dẫn giải:

Đáp án đúng là: A

Ta có $\frac{7}{4}-\left(\frac{x}{60}+\frac{5}{3}\right)=\frac{-12}{5}$ .

 $\begin{array}{l}\frac{x}{60}+\frac{5}{3}=\frac{7}{4}-\left(\frac{-12}{5}\right)\\ \frac{x}{60}+\frac{5}{3}=\frac{7}{4}+\frac{12}{5}\\ \frac{x}{60}+\frac{5}{3}=\frac{35}{20}+\frac{48}{20}\\ \frac{x}{60}+\frac{5}{3}=\frac{83}{20}\\ \frac{x}{60}=\frac{83}{20}-\frac{5}{3}\\ \frac{x}{60}=\frac{249}{60}-\frac{100}{60}\\ \frac{x}{60}=\frac{149}{60}\\ x=\frac{149}{60}.60\end{array}$

x = 149

Vậy ta chọn phương án A.

Câu 3. Kết luận nào đúng về giá trị của biểu thức $A=\frac{1}{5}-\left[\left(\frac{-2}{3}\right)-\left(\frac{1}{3}+\frac{5}{6}\right)\right]?$   

A. A < 2

B. A > 2

C. A < 1

D. A < 0

Hướng dẫn giải

Đáp án đúng là: B

Ta có: $A=\frac{1}{5}-\left[\left(\frac{-2}{3}\right)-\left(\frac{1}{3}+\frac{5}{6}\right)\right]$

 $=\frac{1}{5}-\left[\frac{-\text{ 2}}{3}-\frac{1}{3}-\frac{5}{6}\right]$

 $=\frac{1}{5}-\left[\frac{-\text{ 3}}{3}-\frac{5}{6}\right]$

$=\frac{1}{5}-\left[-1-\frac{5}{6}\right]$

$=\frac{1}{5}-\left[\frac{-6}{6}-\frac{5}{6}\right]$

$=\frac{1}{5}-\left[\frac{-\text{ 11}}{6}\right]$

$=\frac{1}{5}+\frac{11}{6}$

$=\frac{6}{30}+\frac{\text{55}}{30}$

$=\frac{61}{30}.$

Do $A=\frac{61}{30}>\frac{60}{30}=2$  nên A > 2.

Vậy ta chọn phương án B.