Bài 1 (2 điểm). Thực hiện phép tính: 1) A= $\frac{0,5 - \frac{3}{17}}{\frac{5}{6}-\frac{5}{17}} + \frac{0,5-\frac{1}{3}-0,2}{\frac{7}{5}+ \frac{7}{3}-3,5}$ 2) B= $(\frac{1}{121}- \frac{1}{2^2}.( \frac{1}{121}-\frac{1}{3^2})- (\frac{1}{121}-\frac{1}{2022^2})$ 

Đáp án:

a.$A=\dfrac{16}{35}$

b.$B=0$

Giải thích các bước giải:

a.Ta có:

$A=\dfrac{0.5-\dfrac3{17}}{\dfrac56-\dfrac5{17}}+\dfrac{0.5-\dfrac13-0.2}{\dfrac75+\dfrac73-3.5}$

$\to A=\dfrac{\dfrac12-\dfrac3{17}}{\dfrac56-\dfrac5{17}}+\dfrac{\dfrac12-\dfrac13-\dfrac15}{\dfrac75+\dfrac73-\dfrac72}$

$\to A=\dfrac{3(\dfrac16-\dfrac1{17})}{5(\dfrac16-\dfrac1{17})}+\dfrac{\dfrac12-\dfrac13-\dfrac15}{-7(\dfrac12-\dfrac13-\dfrac15)}$

$\to A=\dfrac35-\dfrac17$

$\to A=\dfrac{3\cdot 7-5}{5\cdot 7}$

$\to A=\dfrac{16}{35}$

b.Ta có:

$B=(\dfrac1{121}-\dfrac1{2^2})(\dfrac{121}-\dfrac1{3^2})....(\dfrac1{121}-\dfrac1{2022^2})$

$\to B=(\dfrac1{121}-\dfrac1{2^2})(\dfrac{121}-\dfrac1{3^2})...(\dfrac{121}-\dfrac1{11^2})....(\dfrac1{121}-\dfrac1{2022^2})$

$\to B=(\dfrac1{121}-\dfrac1{2^2})(\dfrac{121}-\dfrac1{3^2})...(\dfrac{121}-\dfrac1{121})....(\dfrac1{121}-\dfrac1{2022^2})$

$\to B=(\dfrac1{121}-\dfrac1{2^2})(\dfrac{121}-\dfrac1{3^2})...0....(\dfrac1{121}-\dfrac1{2022^2})$

$\to B=0$