Đáp án:1a)P=4x+6
b)$x=\frac{-7}{4}$
2a)$\frac{a(a+2)^{2}}{4(a+1)^{2}(a-1)}$
b)P=$\frac{8}{9}$
Giải thích các bước giải:
1)P=$\frac{2}{2x+3}+\frac{3}{2x+1}-\frac{6x+5}{(2x+1)(2x+3)} $
Đk: $x\neq\frac{-3}{2}; x\neq\frac{-1}{2}$
⇒$2(2x+1)+3(2x+3)-6x-5=4x+2+6x+9-6x-5=4x+6$
b)Để $P=-1⇒ 4x+6=-1⇔ 4x=-7⇔ x=\frac{-7}{4}$
2)P=$(\frac{a+1}{2a-2}+\frac{1}{2-2a^{2}}): \frac{2a+2}{a+2}$
=$(\frac{a+1}{2(a-1)}-\frac{1}{2(a+1)(a-1)}):\frac{2(a+1)}{a+2}$
=$(\frac{(a+1)^{2}-1}{2(a-1)(a+1)})·\frac{a+2}{2(a+1)}$
=$\frac{a(a+2)}{2(a+1)(a-1)}·\frac{a+2}{2(a+1)}$
=$\frac{a(a+2)^{2}}{4(a+1)^{2}(a-1)}$
b)khi $|a|=2⇒ a=2$
⇒P=$\frac{2(2+2)^{2}}{4(2+1)^{2}(2-1)}=\frac{32}{38}=\frac{8}{9}$