Ta có:\(\frac{\left[x\left(x-2\right)\right]}{x^2+8x-20}+12x-3=\frac{x\left(x-2\right)}{x^2-2x+10x-20}+12x-3\)

\(=\frac{x\left(x-2\right)}{x\left(x-2\right)+10\left(x-2\right)}+12x-3=\frac{x\left(x-2\right)}{\left(x+10\right)\left(x-2\right)}+12x-3\)

\(=\frac{x}{x+10}+12x-3=\frac{x+\left(12x-3\right).\left(x+10\right)}{x+10}=\frac{x+12x^2+120x-3x-30}{x+10}\)

\(=\frac{12x^2+118x-30}{x+10}\)

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

TH1: \(\left(x-2\right)< 0\)

\(\Rightarrow A=\frac{-x\left(x-2\right)}{x^2+8x-20}=\frac{-x\left(x-2\right)}{x^2-2x+10x-20}=\frac{-x\left(x-2\right)}{\left(x^2-2x\right)+\left(10x-20\right)}\)

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

TH2: \(\left(x-2\right)>0\)

\(\Rightarrow A=\frac{x\left(x-2\right)}{x^2+8x-20}=\frac{x\left(x-2\right)}{x^2-2x+10x-20}=\frac{x\left(x-2\right)}{\left(x^2-2x\right)+\left(10x-20\right)}\)

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

HOK TOT

chệuuu!!

Bài làm

Ta có: A = x| x-2 | / x²+ 8x - 20

A = x| x - 2 | / x² - 2x + 10x - 20

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

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

Nếu x ≥ 2 => x - 2 ≥ 0 => |x - 2| <=> x - 2

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

Nếu x ≤ 2 => x - 2 ≤ 0 => | x - 2 | = -( x - 2 )

Nên A = x.[ -( x - 2 ) ]/ ( x + 10 )( x + 2 ) = -x/ x + 10

Vậy từ biểu thức trên, ta có thể rút gọn thành hai biểu thức mới là A = x/ x + 10 và A = -x/ x +10

Do mik lm bằng đt nên k vt đc phân số. Thông cảm 

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

\(|x-2|=x-2\)khi x-2 >=0 hay x>=2

\(|x-2|=-\left(x-2\right)\)khi x-2<0 hay x<2

Vậy với x>=2 ta có: \(A=\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}=\frac{x}{x+10}\)

với x<2 ta có: \(A=\frac{-x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}=\frac{-x}{x+10}\)

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!

\(A=\)\(\frac{x|x-2|}{x^2+8x-20}+12x-3.\)

\(=\frac{x|x-2|}{\left(x-2\right)\left(x+10\right)}+12x-3\)

Nếu \(x\ge2\Rightarrow x-2\ge0\Leftrightarrow|x-2|=x-2\)

\(\Rightarrow A=\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3=\frac{x}{x+10}+12x-3\)

Nếu \(x< 2\Rightarrow x-2< 0\Leftrightarrow|x-2|=-\left(x-2\right)\)

\(\Rightarrow A=\frac{-x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3=\frac{-x}{x+10}+12x-3\)

Cảm ơn bạn