Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
` @ \color{Red}{m}`
` \color{lightblue}{Answer}`
\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x}{x+1}\)
__
\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\\ =\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\\ =\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\\ =\dfrac{3x-2x+6}{2x\left(x+3\right)}\\ =\dfrac{x+6}{2x\left(x+3\right)}\)
__
\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\\ =\dfrac{1}{1-x}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{1-x}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x-2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1}{1+x}\)
\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)\)
\(=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x+1}\)
========================
\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\)
\(=\dfrac{3x-2\left(x-3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x+6}{2x^2+6x}\)
==========================
\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)\)
\(=-\dfrac{1}{x-1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-\left(x+1\right)+2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-1+2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x+1}\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
\(K=\dfrac{2x^3+x^2-x-2x^3+2-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-x^2}{2x-1}\)
\(=\dfrac{x^2-x+2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{-x\left(x-1\right)}{2x-1}\)
\(=\dfrac{-2x+1}{x^2+x+1}\cdot\dfrac{-x}{2x-1}=\dfrac{x}{x^2+x+1}\)