chịu

\(\dfrac{8-2x}{x^2+x-20}=-\dfrac{2\left(4-x\right)}{\left(4-x\right)\left(x+5\right)}=\dfrac{-2}{x+5}\)

Để biểu thức trên nhận giá trị dương khi 

\(x+5< 0\)do -2 < 0 

\(\Leftrightarrow x< -5\)

\(ĐK:x\ne0\)

Vậy tại x=0 thì k có gt nào của B thỏa mãn

B=x²(x+y)-y(x²-y)+2014

   = x³+x²y-x²y+y²+2014

   = x³+y²+2014

   = 1³+(-1)²+2014

   = 1+1+2014

   =2016

B = x².x+x².y-y.x²+y.y+2014            Uy tín:)

  =  x³+x²y-x²y+y²+2014

  =  x³+y²+2014

Thay x=1;y=-1. Ta có:

B = 1³+(-1)²+2014

   = 1+1+2014

   = 2016

Ta có: \(B=x^2\left(x+y\right)-y\left(x^2-y\right)+2014\)

\(=x^3+x^2y-x^2y-y^2+2014\)

\(=x^3-y^2+2014\)

\(=1-1+2014\)

=2014

a: TXĐ: D=[0;+\(\infty\))\{1}

\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}-\dfrac{\sqrt{x}}{x-1}\)

\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot2}\)

\(=\dfrac{-1}{\sqrt{x}+1}\)

\(a,ĐK:x\ge0\\ x\ne1\\ B=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{2\left(1-\sqrt{x}\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\\ b,x=3\Leftrightarrow B=\dfrac{-1}{\sqrt{3}+1}=\dfrac{1-\sqrt{3}}{2}\\ c,\left|B\right|=\dfrac{1}{2}\Leftrightarrow\left|\dfrac{-1}{\sqrt{x}+1}\right|=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{2}\left(\sqrt{x}+1\ge1>0\right)\\ \Leftrightarrow\sqrt{x}+1=2\Leftrightarrow x=1\left(tm\right)\)

a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=5-1=4\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

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

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

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

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

c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)

d: |B|=A

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

=>\(\dfrac{2}{\sqrt{x}+1}=4\) hoặc \(\dfrac{2}{\sqrt{x}+1}=-4\)

=>\(\sqrt{x}+1=\dfrac{1}{2}\) hoặc \(\sqrt{x}+1=-\dfrac{1}{2}\)

=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)

* Rút gọn biểu thức:

+ Ngoặc thứ nhất:

+ Ngoặc thứ hai:

Do đó:

* Tại  , giá trị biểu thức bằng: 

Ta có:

* Nếu x > 0 thì |x| = x

Ta có: 4x - 8 + |x| = 4x -  8  +x = 5x -  8

Với x = - 2  ta có: 5(- 2 ) - 8 = -5 2  - 2 2  = -7 2

* Nếu -2 < x < 0 thì |x| = -x

Ta có: 4x -  8  + |x| = 4x -  8  - x = 3x -  8

Với x = - 2  ta có: 3(- 2  ) -  8  = -3 2  - 2 2  = -5 2

`A=(x^2-x-6)/x+2`

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

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

`x=-2`

`=>A=(4+2-6)/(-2)`

`=0/(-2)`

`=0`

Ta có: \(A=\dfrac{x^2-x-6}{x+2}\)

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

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

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

=x-3

Vì x=-2 không thỏa mãn ĐKXĐ nên Khi x=-2 thì \(A\in\varnothing\)