Tính: (a+1+ 1-2a^2/a-1):(1-1/1-a)

Giải SBT Toán 8 Bài tập cuối chương 1 trang 26

Bài 15 trang 27 SBT Toán 8 Tập 1: 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:

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

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

$=\frac{a^2-1+1-2a^2}{a-1}:\frac{-a}{1-a}$

$=\frac{-a^2}{a-1}.\frac{1-a}{-a}$

$=\frac{-a^2}{a-1}.\frac{a-1}{a}=-a$.

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

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

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

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

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

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

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

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

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

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

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

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