Question

Cho biểu thức: \(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)với \(x\ne0,x\ne-1,x\ne-2,x\ne2\)

a, Rút gọn A

b, Tính A khi \(x\)thỏa mãn \(^{x^2-2x=8}\)

Answer 1

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 ) 

Answer 2

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

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

Answer 3

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

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

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

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

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

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

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

Answer 4

b,Ta có:\(x^2-2x=8\)

\(\Leftrightarrow x^2-2x-8=0\)

\(\Leftrightarrow x^2+2x-4x-8=0\)

\(\Leftrightarrow x\left(x+2\right)-4\left(x+2\right)=9=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\left(ktmđkxđ\right)\\x=4\left(tmđkxđ\right)\end{cases}}}\)

Với \(x=4\Rightarrow A=\frac{1}{4+1}=\frac{1}{5}\)