`a)M=(x^4+2)/(x^6+1)+(x^2-1)/(x^4-x^2+1)-(x^2+3)/(x^4+4x^2+3)`

`=(x^4+2)/(x^6+1)+(x^2-1)/(x^4-x^2+1)-(x^2+3)/((x^2+1)(x^2+3))`

`=(x^4+2)/(x^6+1)+((x^2-1)(x^2+1))/(x^6+1)-1/(x^2+1)`

`=(x^4+2+x^4-1-x^4+x^2-1)/(x^2+1)`

`=(x^4+x^2)/(x^2+1)`

`=(x^2(x^2+1))/(x^2+1)`

`=x^2`

`b)` tìm gtnn chứ?

`M=x^2>=0`

Dấu '=" `<=>x=0`

\(A=\dfrac{2}{x-1}\sqrt{\dfrac{\left(x-1\right)^2}{4x^2}}=\dfrac{2}{x-1}\left|\dfrac{x-1}{2x}\right|=\dfrac{\left|x-1\right|}{\left(x-1\right)\left|x\right|}\)

\(B=\left(x^2-4\right)\sqrt{\dfrac{9}{x^2-4x+4}}=\dfrac{3\left(x^2-4\right)}{\left|x-2\right|}\)

a) Ta có: \(A=\dfrac{2}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)

\(=\dfrac{2}{x-1}\cdot\dfrac{x-1}{2x}\)

\(=\dfrac{1}{x}\)

b) Ta có: \(\left(x^2-4\right)\cdot\sqrt{\dfrac{9}{x^2-4x+4}}\)

\(=\dfrac{\left(x-2\right)\left(x+2\right)\cdot3}{\left(x-2\right)^2}\)

\(=\dfrac{3x+6}{x-2}\)

1:

a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

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

\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)

\(=\left(-3\right)^2+5\left(2x-3\right)\)

\(=9+10x-15=10x-6\)

2: 

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

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

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

ĐKXĐ: \(x\ne\pm1\)

\(A=\left(\dfrac{\left(1+x\right)^2}{\left(1-x\right)\left(1+x\right)}-\dfrac{\left(1-x\right)^2}{\left(1-x\right)\left(1+x\right)}+\dfrac{4x^2}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)

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

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

\(=\dfrac{4x\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)}{4\left(x+1\right)}=-\dfrac{x}{x+1}\)

\(A=\frac{2x^2+4x}{x^3-4x}+\frac{x^2-4}{x^2+2x}+\frac{2}{2-x}\left(x\ne0;x\ne\pm2\right)\)

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

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

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

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

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

Vậy \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)

b) \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)

Ta có: x=4 (tmđk) thay vào A ta có:

\(A=\frac{-2\cdot4+8}{4\left(4-2\right)}=\frac{-8+8}{4\cdot2}=\frac{0}{8}=0\)

Vậy A=0 với x=4

a: \(A=\left(\dfrac{4}{x}-1\right):\left(1-\dfrac{x-3}{x^2+x+1}\right)\)

\(=\dfrac{4-x}{x}:\dfrac{x^2+x+1-x+3}{x^2+x+1}\)

\(=\dfrac{4-x}{x}\cdot\dfrac{x^2+x+1}{x^2+4}=\dfrac{\left(4-x\right)\left(x^2+x+1\right)}{x\left(x^2+4\right)}\)

b: x^4-7x^2-4x+20=0

=>(x-2)^2(x^2+4x+5)=0

=>x=2

Khi x=2 thì \(A=\dfrac{\left(4-2\right)\left(4+2+1\right)}{2\left(4+4\right)}=\dfrac{7}{8}\)

a: \(A=\left[\left(\dfrac{4x}{x+2}+\dfrac{8x^2}{4-x^2}\right)\right]:\left[\dfrac{x-1}{x^2-2x}-\dfrac{2}{x}\right]\)

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

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

\(=\dfrac{-8x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-2\right)}{x-1-2x+4}\)

\(=\dfrac{-8x^2}{\left(x+2\right)\cdot\left(-x+3\right)}\)

\(=\dfrac{8x^2}{\left(x-3\right)\left(x+2\right)}\)

b: \(x^2+2x=15\)

=>\(x^2+2x-15=0\)

=>(x+5)(x-3)=0

=>\(\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Thay x=-5 vào A, ta được:

\(A=\dfrac{8\cdot\left(-5\right)^2}{\left(-5-3\right)\left(-5+2\right)}=\dfrac{8\cdot25}{\left(-8\right)\cdot\left(-3\right)}=\dfrac{25}{3}\)

c: |A|>A

=>A<0

=>\(\dfrac{8x^2}{\left(x-3\right)\left(x+2\right)}< 0\)

=>(x-3)(x+2)<0

TH1: \(\left\{{}\begin{matrix}x-3>0\\x+2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 3\\x>-2\end{matrix}\right.\)

=>-2<x<3

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}-2< x< 3\\x\notin\left\{0;2\right\}\end{matrix}\right.\)