Ta có : \(x^2+8x-20=\left(x-2\right)\left(x+10\right)\)

\(\left|x-2\right|=x-2\Leftrightarrow x\ge0\)

\(\left|x-2\right|=-\left(x-2\right)\Leftrightarrow x\le0\)

Vì \(x\ge0\)suy ra : \(\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}=\frac{x}{x+10}\)

Vì \(x\le0\)suy ra : \(\frac{x\left[-\left(x-2\right)\right]}{\left(x-2\right)\left(x+10\right)}=\frac{-x}{x+10}\)

x² + 8x - 20 = x² + 10x - 2x - 20 = x(x+10) - 2(x+10) = (x - 2)(x+ 10)

|x - 2| = x - 2 nếu x > 2; |x - 2| = -(x - 2) = - x + 2 nếu x < 2

Vậy A = \(\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}=\frac{x}{x+10}\) nếu x > 2

A = \(\frac{-x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}=\frac{-x}{x+10}\) nếu x < 2

A=(x/x-2/)/x^2+8x-20=(x/x-2/)/(x-2).(x+10)

TH1:x>=2

A=x.(x-2)/(x-2).(x+10)=x/x+10

TH2:x<2

A=(-x).(x-2)/(x-2).(x+10)=-x/x+10

bài của tôi dể hiểu hơn.Hãy k cho tôi!

Để \(\sqrt{x}\) xác định

 \(\Leftrightarrow x\ge0\)

\(\Leftrightarrow-7x\le0\)

\(\Rightarrow\sqrt{-7x}\)không tồn tại 

\(\Leftrightarrow\frac{8x}{4x\sqrt{x-8x}}\)không tồn tại

=> A không tồn tại 

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

b) \(\frac{x^2-16}{2x^2-8x}=\frac{\left(x-4\right)\left(x+4\right)}{2x\left(x-4\right)}=\frac{x+4}{2x}\)

\(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x^2\right)+17\)

\(=5\left(x^2-4\right)-\left(3-4x^2\right)+17\)

\(=5x^2-20-3+4x^2+17\)

\(=9x^2-6\)