a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

Bài 1: 

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

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

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

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

chịu

B = (x – 2)( x 2 + 2x + 4) – x(x – 1)(x + 1) + 3x

B = (x – 2)( x 2 + x.2 + 2 2 ) – x( x 2 – 1) + 3x

B   =   x 3   –   2 3   –   x . x 2   +   x . 1   +   3 x     B   =   x 3   –   8   –   x 3   +   x   +   3 x

B = 4x – 8

Đáp án cần chọn là: D

Ta có: \(B=\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{4}{1-x^2}\)

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

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

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

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

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

\(\dfrac{x-1}{x+1}\) - \(\dfrac{x+1}{-\left(1-x\right)}\) - \(\dfrac{4}{\left(1-x\right)\left(1+x\right)}\)  MTC: -(1 - x)(1 + x)

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

= ​​\(\dfrac{-x^2+2x-1}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{x^2+2x+1}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{-4}{\text{-(1 - x)(1 + x)}}\)

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

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a: Đặt A=|x-2|+|2x-1|

TH1: x<1/2

=>2x-1<0 và x-2<0

A=|x-2|+|2x-1|

=2-x+1-2x

=-3x+3

TH2: 1/2<=x<2

=>2x-1>=0 và x-2<0

=>A=2-x+2x-1=x+1

TH3: x>=2

=>2x-1>0 và x-2>=0

=>A=2x-1+x-2=3x-3

b: Đặt B=|4-3x|-|2x+1|

=|3x-4|-|2x+1|

TH1: x<-1/2

=>\(2x+1< 0;3x-4< 0\)

=>\(B=4-3x-\left(-2x-1\right)\)

\(=4-3x+2x+1\)

\(=5-x\)

TH2: \(-\dfrac{1}{2}< =x< \dfrac{4}{3}\)

=>\(2x+1>=0;3x-4< 0\)

=>\(B=4-3x-\left(2x+1\right)\)

\(=4-3x-2x-1=-5x+3\)

TH3: \(x>=\dfrac{4}{3}\)

=>\(3x-4>=0;2x+1>0\)

=>\(B=3x-4-\left(2x+1\right)\)

\(=3x-4-2x-1\)

=x-5

a: ĐKXĐ: \(x\notin\left\{-1;1;0\right\}\)