Bài 6 trang 25 Toán lớp 7: Tính nhanh:

a)$\frac{13}{23}.\frac{7}{11}+\frac{10}{23}.\frac{7}{11};$                                          

b) $\frac{5}{9}.\frac{23}{11}-\frac{1}{11}.\frac{5}{9}+\frac{5}{9}$

c)$\left[\left(-\frac{4}{9}\right)+\frac{3}{5}\right]:\frac{13}{17}+\left(\frac{2}{5}-\frac{5}{9}\right):\frac{13}{17};$                

d) $\frac{3}{16}:\left(\frac{3}{22}-\frac{3}{11}\right)+\frac{3}{16}:\left(\frac{1}{10}-\frac{2}{5}\right)$

Phương pháp giải:

Lời giải:

a)

$\begin{array}{l}\frac{13}{23}.\frac{7}{11}+\frac{10}{23}.\frac{7}{11}\\ =\frac{7}{11}\left(\frac{13}{23}+\frac{10}{23}\right)\\ =\frac{7}{11}.1\\ =\frac{7}{11}\end{array}$                                   

b)

$\begin{array}{l}\frac{5}{9}.\frac{23}{11}-\frac{1}{11}.\frac{5}{9}+\frac{5}{9}\\ =\frac{5}{9}.\left(\frac{23}{11}-\frac{1}{11}+1\right)\\ =\frac{5}{9}.\left(2+1\right)\\ =\frac{5}{9}.3=\frac{5}{3}\end{array}$

c)

$\begin{array}{l}\left[\left(-\frac{4}{9}+\frac{3}{5}\right):\frac{13}{17}\right]+\left(\frac{2}{5}-\frac{5}{9}\right):\frac{13}{17}\\ =\left(-\frac{4}{9}+\frac{3}{5}\right).\frac{17}{13}+\left(\frac{2}{5}-\frac{5}{9}\right).\frac{17}{13}\\ =\frac{17}{13}.\left(-\frac{4}{9}+\frac{3}{5}+\frac{2}{5}-\frac{5}{9}\right)\\ =\frac{17}{13}.\left[\left(-\frac{4}{9}-\frac{5}{9}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\right]\\ =\frac{17}{13}.\left(-1+1\right)\\ =\frac{17}{13}.0=0\end{array}$          

d)

$\begin{array}{l}\frac{3}{16}:\left(\frac{3}{22}-\frac{3}{11}\right)+\frac{3}{16}:\left(\frac{1}{10}-\frac{2}{5}\right)\\ =\frac{3}{16}:\left(\frac{3}{22}-\frac{6}{22}\right)+\frac{3}{16}:\left(\frac{1}{10}-\frac{4}{10}\right)\\ =\frac{3}{16}:\frac{-3}{22}+\frac{3}{16}:\frac{-3}{10}\\ =\frac{3}{16}.\frac{-22}{3}+\frac{3}{16}.\frac{-10}{3}\\ =\frac{3}{16}.\left(\frac{-22}{3}+\frac{-10}{3}\right)\\ =\frac{3}{16}.\frac{-32}{3}\\ =-2\end{array}$

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.