cái này nó hơi khó 1 tí nên chú ý chút khác lên lever :>

a, \(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)ĐK : x khác 0 ; 2 ; -2

\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)

\(=\left(\frac{4x\left(x-2\right)}{MTC}+\frac{2x\left(x+2\right)}{MTC}+\frac{\left(6-5x\right)x}{MTC}\right):\frac{x+1}{x-2}\)

\(=\left(\frac{4x^2-8x+2x^2+4x+6x-5x^2}{MTC}\right):\frac{x+1}{x-2}\)

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

b, Ta có : \(x^2-2x=8\Leftrightarrow x^2-2x-8=0\)

\(\left(x-4\right)\left(x+2\right)=0\)<=> \(x=4;-2\)

TH1 : Thay x = 4 ta được : \(\frac{1}{4+1}=\frac{1}{5}\)

TH2 : Thay x = -2 ta được : ( ktmđkxđ ) 

\(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right)\div\frac{x+1}{x-2}\)

a)\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\right)\times\frac{x-2}{x+1}\)

\(=\left(\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)

\(=\left(\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)

\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}\times\frac{x-2}{x+1}\)

\(=\frac{1}{x+1}\)

b) x² - 2x = 8

<=> x² - 2x - 8 = 0

<=> x² - 4x + 2x - 8 = 0

<=> x( x - 4 ) + 2( x - 4 ) = 0

<=> ( x - 4 )( x + 2 ) = 0

<=> x = 4 ( tm ) hoặc x = -2 ( ktm )

Với x = 4 ( tm ) => A = 1/5

Với x = -2 ( ktm ) => A không xác định

a,\(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right)\div\frac{x+1}{x-2}\)

\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right)\div\frac{x+1}{x-2}\)

\(=\left(\frac{4x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{x\left(6-5x\right)}{x\left(x-2\right)\left(x+2\right)}\right)\div\frac{x+1}{x-2}\)

\(=\frac{4x^2-8x+2x^2+4x+6x-5x^2}{x\left(x-2\right)\left(x+2\right)}\div\frac{x+1}{x-2}\)

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

\(=\frac{x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\div\frac{x+1}{x-2}\)

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

\(C=\left[\frac{x^2.\left(x^2-4\right)+4x^2}{x^2-4}\right].\left[\frac{x}{2.\left(x-2\right)}+\frac{2-2x}{x.\left(x^2-4\right)}.\frac{x^2-4}{x-2}\right]\)

\(C=\frac{x^4-4x^2+4x^2}{x^2-4}.\left[\frac{x}{2.\left(x-2\right)}+\frac{2-2x}{x\left(x-2\right)}\right]\)

\(C=\frac{x^4}{x^2-4}.\left[\frac{x^2}{2x.\left(x-2\right)}+\frac{\left(2-2x\right).2}{2x.\left(x-2\right)}\right]\)

\(C=\frac{x^4}{x^2-4}.\left[\frac{x^2+4-4x}{2x.\left(x-2\right)}\right]\)

\(C=\frac{x^4}{x^2-4}.\frac{\left(x-2\right)^2}{2x.\left(x-2\right)}\)

\(C=\frac{x^4}{\left(x-2\right).\left(x+2\right)}.\frac{\left(x-2\right).\left(x-2\right)}{2x.\left(x-2\right)}\)

\(C=\frac{x^3}{\left(x+2\right).2}\)

Bạn chưa rút gọn hết, Despacito.

a)  \(ĐKXĐ:x\ne\pm2\)

\(P=\left[\frac{x^2+2x}{x^3+2x^2+4x+8}+\frac{2}{x^2+4}\right]:\left[\frac{1}{x-2}-\frac{4x}{x^3-2x^2+4x-8}\right]\)

\(\Leftrightarrow P=\left(\frac{x}{x^2+4}+\frac{2}{x^2+4}\right):\left(\frac{1}{x-2}-\frac{4x}{\left(x-2\right)\left(x^2+4\right)}\right)\)

\(\Leftrightarrow P=\frac{x+2}{x^2+4}:\frac{x^2+4-4x}{\left(x-2\right)\left(x^2+4\right)}\)

\(\Leftrightarrow P=\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x^2+4\right)\left(x-2\right)^2}\)

\(\Leftrightarrow P=\frac{x+2}{x-2}\)

b) P là số nguyên tố khi và chỉ khi \(x+2⋮x-2\)

\(\Leftrightarrow4⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow x\in\left\{1;3;0;4;-2;6\right\}\)

Loại \(x=-2\)

\(\Leftrightarrow P\in\left\{-3;5;-1;3;2\right\}\)

Vì P là số nguyên tố nên

\(P\in\left\{5;3;2\right\}\)

Vậy để P là số nguyên tố thì  \(x\in\left\{3;4;6\right\}\)

Bài 1.

a)\(\frac{4x-4}{x^2-4x+4}\div\frac{x^2-1}{\left(2-x\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\div\frac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\times\frac{\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4}{x+1}\)

b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)

\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)

c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)

\(=\left(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x}{x^2-1}\right)\times\frac{x-1}{4x}\)

\(=\left(\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)

\(=\left(\frac{x-1+x+1+2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)

\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)

Bài 3.

N = ( 4x + 3 )² - 2x( x + 6 ) - 5( x - 2 )( x + 2 )

= 16x² + 24x + 9 - 2x² - 12x - 5( x² - 4 )

= 14x² + 12x + 9 - 5x² + 20

= 9x² + 12x + 29

= 9( x² + 4/3x + 4/9 ) + 25

= 9( x + 2/3 )² + 25 ≥ 25 > 0 ∀ x 

=> đpcm