Question

Упростите выражения. #736

Answer 1

$1)\;\frac{9a^2-4}{2a^2-5x+2}\cdot\frac{a-2}{3a+2}+\frac{a-1}{1-2a}=\frac{(3a-2)(3a+2)}{(a-2)(2a-1)}\cdot\frac{a-2}{3a+2}+\frac{a-1}{1-2a}=\frac{3a-2}{2a-1}-\frac{a-1}{2a-1}=\\\\=\frac{3a-2-a+1}{2a-1}=\frac{2a-1}{2a-1}=1$

$2)\;\frac{b-4}{b^3-b}:\left(\frac{b-1}{2b^2+3b+1}-\frac1{b^2-1}\right)=\frac{b-4}{b(b^2-1)}:\left(\frac{b-1}{(2b+1)(b+1)}-\frac1{(b-1)(b+1)}\right)=\\=\frac{b-4}{b(b^2-1)}:\frac{(b-1)^2-(2b+1)}{(2b+1)(b+1)(b-1)}=\frac{b-4}{b(b^2-1)}\cdot\frac{(2b+1)(b^2-1)}{b^2-2b+1-2b-1}=\frac{b-4}b\cdot\frac{2b+1}{b^2-4b}=\\\\=\frac{b-4}b\cdot\frac{2b+1}{b(b-4)}=\frac{2b+1}b$

$3)\;\left(\frac{c+2}{c^2-c-6}-\frac{2c}{c^2-6c+9}\right):\frac{c^2+3c}{(2c-6)^2}=\left(\frac{c+2}{(c-3)(c+2)}-\frac{2c}{(c-3)^2}\right)\cdot\frac{(2c-6)^2}{c^2+3c}=\\\\=\left(\frac1{c-3}-\frac{2c}{(c-3)^2}\right)\cdot\frac{4(c-3)}{c(c+3)}=\frac{c-3-2c}{(c-3)^2}\cdot\frac{4(c-3)}{c(c+3)}=\frac{-(c+3)}{(c-3)}\cdot\frac4{c(c+3)}=-\frac4{c(c-3)}$