\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)

\(=\left(\dfrac{x}{x+1}.\dfrac{x^2+x+1}{x^2}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)=\left(\dfrac{x^2+x-1}{x^2+x}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)\)

\(=\left(\dfrac{x^2+x-1-2x-1}{x\left(x+1\right)}\right)\left(x+1\right)=\dfrac{x^2-3x-2}{x}\)

à xin lỗi mình nhầm dòng cuối 

\(=\dfrac{x^2-x-2}{x}=\dfrac{\left(x+1\right)\left(x-2\right)}{x}\)

Để biểu thức trên nhận giá trị dương khi 

\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi 

a) Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

\(=x^3+1-x^3+1\)

=2

b) Ta có: \(\left(2x+3\right)\left(2x-3\right)-\left(2x+1\right)^2\)

\(=4x^2-9-4x^2-4x-1\)

\(=-4x-10\)

\(P=\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3-x^2+x-1}\right):\left(\dfrac{1-2x}{x+1}\right)\left(ĐKXĐ:x\ne0;x\ne\pm1\right)\)

\(=\left(\dfrac{1}{x-1}-\dfrac{2x}{x^2\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{1-2x}{x+1}\right)\)

\(=\left(\dfrac{1}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x^2+1\right)}\right):\left(\dfrac{1-2x}{x+1}\right)\)

\(=\left(\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\right):\left(\dfrac{1-2x}{x+1}\right)\)

\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}:\dfrac{1-2x}{x+1}\)

\(=\dfrac{x-1}{x^2+1}:\dfrac{1-2x}{x+1}\)

\(=\dfrac{x-1}{x^2+1}.\dfrac{x+1}{1-2x}\)

\(=\dfrac{x^2-1}{\left(x^2+1\right)\left(1-2x\right)}\)

ĐKXĐ: \(x\notin\left\{1;\dfrac{1}{2}\right\}\)

\(\left(\dfrac{1}{x-1}+2+\dfrac{2x^3+x^2-x}{1-x^3}\right):\dfrac{1-2x}{x^3+x-2}\)

\(=\left(\dfrac{1}{x-1}+2-\dfrac{2x^3+x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right)\cdot\dfrac{x^3+x-2}{1-2x}\)

\(=\dfrac{x^2+x+1+2\left(x^3-1\right)-2x^3-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^3-x^2+x^2-x+2x-2}{-\left(2x-1\right)}\)

\(=\dfrac{2x+1+2x^3-2-2x^3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+2\right)}{-\left(2x-1\right)}\)

\(=\dfrac{2x-1}{x^2+x+1}\cdot\dfrac{-\left(x^2+x+2\right)}{2x-1}=\dfrac{-x^2-x-2}{x^2+x+1}\)

có phải ý bạn là:

rút gọn biểu thức:

\(\frac{x}{\left(x+1\right)^3}\cdot\frac{1}{x+1}+\frac{1}{x^2+2x+1}\cdot\frac{1}{x^2+1}:\frac{x-1}{x^3}\)

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

f: Ta có: \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)

\(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)

\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)

\(=x^3-16x^2+25x\)

g: Ta có: \(3\left(x+2\right)^2-\left(3x+1\right)\left(x+5\right)+\left(x+5\right)^2\)

\(=3\left(x^2+4x+4\right)-\left(3x^2+16x+5\right)+x^2+10x+25\)

\(=3x^2+12x+12-3x^2-16x-5+x^2+10x+25\)

\(=x^2+6x+32\)

e) (x+1)²+(x-1)²-2(1+x)(1-x)

= (x+1)² + 2(1+x)(x-1) + (x-1)²

= (x+1+x-1)²

= 4x²

a: Ta có: \(\left(x+1\right)^2+\left(x-1\right)^2-2\left(1+x\right)\left(1-x\right)\)

\(=\left(x+1\right)^2+2\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\)

\(=\left(x+1+x-1\right)^2\)

\(=4x^2\)

c: Ta có: \(3\left(x+2\right)^2-\left(3x+1\right)\left(x+5\right)+\left(x+5\right)^2\)

\(=3x^2+12x+12-3x^2-16x-5+x^2+10x+25\)

\(=x^2+6x+32\)

a: \(=\dfrac{\left(x+1\right)\left[\left(3x-2\right)-\left(2x+5\right)\left(x-1\right)\right]}{x+1}\)

=3x-2-2x^2+2x-5x+5

=-2x^2+3

b: \(=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\)

c: =x^3-3x^2+3x-1-x^3-1+9x^2-1

=6x^2+3x-3

\(a,\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]:\left(x+1\right)\)

\(=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x-1\right)\left(x+1\right)\right]:\left(x+1\right)\)

\(=\left[\left(x+1\right)\left(3x-2-\left(2x+5\left(x-1\right)\right)\right)\right]:\left(x+1\right)\)

\(=\left[\left(x+1\right)\left(3x-2-2x^2+2x-5x+5\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(-2x^2+3\right)\right].\dfrac{1}{x+1}\)

\(=-2x^2+3\)

\(b,\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)\)

\(=\left(2x+1\right)\left[\left(2x+1\right)-2\left(3-x\right)\right]\)

\(=\left(2x+1\right)\left(2x+1-6+2x\right)\)

\(=\left(2x+1\right)\left(4x-5\right)\)

\(c,\left(x-1\right)^3-\left(x+1\right)\left(x^2-x+1\right)-\left(3x+1\right)\left(1-3x\right)\)

\(=x^3-3x^2+3x-1-x^3-1-\left(3x-9x^2+1-3x\right)\)

\(=-3x^2+3x-2-3x+9x^2-1+3x\)

\(=6x^2+3x-3\)