bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a) đk: x khác 0;2;-2;3

A = \(\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)

= \(\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2x-x^2}\)

= \(\left(\dfrac{\left(x+2\right)^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\right):\dfrac{x-3}{x\left(2-x\right)}\)

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

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

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

b) Có \(\left|x-5\right|=2\)

<=> \(\left[{}\begin{matrix}x-5=2< =>x=7\left(Tm\right)\\x-5=-2< =>x=3\left(L\right)\end{matrix}\right.\)

Thay x = 7 vào A, ta có:

\(A=\dfrac{4.7^2}{7-3}=49\)

c) A = \(\dfrac{4x^2}{x-3}⋮4\left(\forall x\right)\)

a) \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\) (ĐK: \(x\ne\pm3\))

\(A=\left[\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\right]:\left(2+\dfrac{x+5}{x+3}\right)\)

\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}:\dfrac{2\left(x+3\right)-\left(x+5\right)}{x+3}\)

\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)

\(A=\dfrac{-5\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}\)

\(A=\dfrac{-5}{x-3}\)

b) Ta có: \(\left|x\right|=1\)

TH1: \(\left|x\right|=-x\) với \(x< 0\)

Pt trở thành:

\(-x=1\) (ĐK: \(x< 0\)) 

\(\Leftrightarrow x=-1\left(tm\right)\)

Thay \(x=-1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{-1-3}=\dfrac{5}{4}\)

TH2: \(\left|x\right|=x\) với \(x\ge0\)

Pt trở thành:

\(x=1\left(tm\right)\) (ĐK: \(x\ge0\)) 

Thay \(x=1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{1-2}=\dfrac{5}{2}\)

c) \(A=\dfrac{1}{2}\) khi:

\(\dfrac{-5}{x-3}=\dfrac{1}{2}\)

\(\Leftrightarrow-10=x-3\)

\(\Leftrightarrow x=-10+3\)

\(\Leftrightarrow x=-7\left(tm\right)\)

d) \(A\) nguyên khi:

\(\dfrac{-5}{x-3}\) nguyên

\(\Rightarrow x-3\inƯ\left(-5\right)\)

\(\Rightarrow x\in\left\{8;-2;2;4\right\}\)

a: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)

\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)

\(=\dfrac{-5x-5}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-5}{x-3}\)

b: |x|=1

=>x=-1(loại) hoặc x=1(nhận)

Khi x=1 thì \(A=\dfrac{-5}{1-3}=-\dfrac{5}{-2}=\dfrac{5}{2}\)

c: A=1/2

=>x-3=-10

=>x=-7

d: A nguyên

=>-5 chia hết cho x-3

=>x-3 thuộc {1;-1;5;-5}

=>x thuộc {4;2;8;-2}

a) ĐKXĐ: \(x\ne-2\)

b) Ta có: \(\dfrac{2x^2-4x+8}{x^3+8}\)

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

\(=\dfrac{2}{x+2}\)

c) Vì x=2 thỏa mãn ĐKXĐ

nên Thay x=2 vào biểu thức \(\dfrac{2}{x+2}\), ta được:

\(\dfrac{2}{2+2}=\dfrac{2}{4}=\dfrac{1}{2}\)

Vậy: Khi x=2 thì giá trị của biểu thức là \(\dfrac{1}{2}\)

d) Để \(\dfrac{2}{x+2}=2\) thì x+2=1

hay x=-1(nhận)

Vậy: Để \(\dfrac{2}{x+2}=2\) thì x=-1

nhanh giùm mình được không

Bài 1: 

a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

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

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)

\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

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

\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(a,A=\dfrac{x^4-5x^2+4}{x^4-x^2+4x-4}=\dfrac{x^4-x^2-4x^2+4}{x^2\left(x-1\right)\left(x+1\right)+4\left(x-1\right)}\\ A=\dfrac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{\left(x-1\right)\left(x^2+x+4\right)}\\ A=\dfrac{\left(x-1\right)\left(x+1\right)\left(x^2-4\right)}{\left(x-1\right)\left(x^2+x+4\right)}=\dfrac{\left(x+1\right)\left(x^2-4\right)}{x^2+x+4}=\dfrac{x^3+x^2-4x-4}{x^2+x+4}\)

`1/2`

`1/2

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

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

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

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

b) Thay \(x=-\dfrac{1}{2}\) vào A, ta được:

\(A=\left(2\cdot\dfrac{1}{4}+1\right):\left(\dfrac{-1}{2}-1\right)\)

\(=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)

c) Để A<1 thì A-1<0

\(\Leftrightarrow\dfrac{2x^2+1}{x-1}-1< 0\)

\(\Leftrightarrow\dfrac{2x^2+1-x+1}{x-1}< 0\)

\(\Leftrightarrow\dfrac{2x^2-x+2}{x-1}< 0\)

\(\Leftrightarrow x-1< 0\)

hay x<1