bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a: \(A=\left(\frac{x}{x+2}+\frac{2}{x-2}+\frac{4x}{4-x^2}\right):\frac{2x+1}{8x+16}\)

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

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

\(=\frac{8\left(x-2\right)}{2x+1}=\frac{8x-16}{2x+1}\)

b: Thay \(x=-2\frac12=-2,5\) vào A, ta được:

\(A=\frac{8\cdot\left(-2,5\right)-16}{2\cdot\left(-2,5\right)+1}=\frac{-20-16}{-5+1}=\frac{-36}{-4}=9\)

c: Để A nguyên thì 8x-16⋮2x+1

=>8x+4-20⋮2x+1

=>-20⋮2x+1

=>2x+1∈{1;-1;5;-5}

=>2x∈{0;-2;4;-6}

=>x∈{0;-1;2;-3}

Kết hợp ĐKXĐ, ta được: x∈{0;-1;-3}

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

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

\(P=A:B=\dfrac{x+2}{x-1}\cdot\dfrac{x+1}{x+2}=\dfrac{x+1}{x-1}\)

3: Để P>1/3 thì \(P-\dfrac{1}{3}>0\)

=>\(\Leftrightarrow3\left(x+1\right)-x+1>0\)

=>3x+3-x+1>0

=>2x+4>0

hay x>-2

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

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

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

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

dell hiểu đc đề ntn cả?

ngu nên đéo hiểu

a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.

Thay x=-2 và B ta có :

\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)

b) Rút gọn : 

\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)

\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

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

Xấu nhỉ ??