Đề bị lỗi công thức rồi bạn.

a, ĐKXĐ: \(x\ne0;x\ne\pm1\)

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

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

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

b,Để \(P=2\Leftrightarrow\frac{x}{x-1}=2\Leftrightarrow2\left(x-1\right)=x\Leftrightarrow2x-2-x=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(tmđk\right)\)

Vậy để P=2 <=> x=2

a) điều kiện xác định: x≠3 và x≠2

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

Tại x=13 ta có \(\dfrac{13+2}{13-3}\)=\(\dfrac{3}{2}\)

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

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mik ko hỉu 

sorry, mn k giùm mik nhé

thanks

 Thực hiện phép tính (10x^5y^2-6x^2y^5+8x^2y^5):(-2x^2y^2)

\(A=\left(x+2\right).\left(x^2-2x+4\right)-\left(18+x^3\right)\)

\(=x^3+8-18-x^3\)

\(=-10\)

\(B=8m-\left(m+3\right)^2+\left(m-3\right).\left(3+m\right)\)

\(=8m-\left(m^2+6m+9\right)+m^2-3^2\)

\(=8m-m^2-6m-9+m^2-9\)

\(=2m-18\)