Giải bài 15 trang 27 sách bài tập toán 8 - Chân trời sáng tạo

Đề bài

Tính:

a) $\left(a+1+\frac{1-2a^2}{a-1}\right):\left(1-\frac{1}{1-a}\right)$;

b) $\left(\frac{a}{b^2}-\frac{1}{a}\right):\left(\frac{1}{b}+\frac{1}{a}\right)$;

c) $\left(a-\frac{4ab}{a+b}+b\right).\left(a+\frac{4ab}{a-b}-b\right)$;

d) $ab+\frac{ab}{a+b}\left(\frac{a+b}{a-b}-a-b\right)$.

Lời giải chi tiết

a) $\left(a+1+\frac{1-2a^2}{a-1}\right):\left(1-\frac{1}{1-a}\right)=\frac{\left(a+1\right)\left(a-1\right)+1-2a^2}{a-1}:\frac{1-a-1}{1-a}$

$=\frac{a^2-1+1-2a^2}{a-1}.\frac{a-1}{a}=\frac{-a^2\left(a-1\right)}{a\left(a-1\right)}=-a$

b) $\left(\frac{a}{b^2}-\frac{1}{a}\right):\left(\frac{1}{b}+\frac{1}{a}\right)=\frac{a^2-b^2}{a{b}^2}:\frac{a+b}{ab}=\frac{\left(a-b\right)\left(a+b\right)ab}{a{b}^2\left(a+b\right)}=\frac{a-b}{b}$

c) $\left(a-\frac{4ab}{a+b}+b\right).\left(a+\frac{4ab}{a-b}-b\right)=\frac{\left(a+b\right)\left(a+b\right)-4ab}{a+b}.\frac{\left(a-b\right)\left(a-b\right)+4ab}{a-b}$

$=\frac{a^2+2ab+b^2-4ab}{a+b}.\frac{a^2-2ab+b^2+4ab}{a-b}=\frac{a^2-2ab+b^2}{a+b}.\frac{a^2+2ab+b^2}{a-b}$

$=\frac{{\left(a-b\right)}^2{\left(a+b\right)}^2}{\left(a+b\right)\left(a-b\right)}=\left(a+b\right)\left(a-b\right)=a^2-b^2$

d) $ab+\frac{ab}{a+b}\left(\frac{a+b}{a-b}-a-b\right)=ab+\frac{ab}{a+b}.\frac{a+b-\left(a-b\right)\left(a+b\right)}{a-b}$

$=ab+\frac{ab}{a+b}.\frac{\left(a+b\right)\left(1-a+b\right)}{a-b}=ab+\frac{ab\left(1-a+b\right)}{a-b}=\frac{ab\left(a-b\right)+ab-a^{2}b+a{b}^2}{a-b}$

$=\frac{a^{2}b-a{b}^2+ab-a^{2}b+a{b}^2}{a-b}=\frac{\left(a^{2}b-a^{2}b\right)+\left(a{b}^2-a{b}^2\right)+ab}{a-b}=\frac{ab}{a-b}$