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

Bài 1 : 

\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)

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

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)

\(\dfrac{20x\left(2-x\right)}{12x\left(x-2\right)^2}=\dfrac{5.4.x\left(2-x\right)}{3.4.x\left(2-x\right)^2}=\dfrac{5}{3\left(2-x\right)}\)

\(=\dfrac{-20x\left(x-2\right)}{12x\left(x-2\right)^2}=\dfrac{-5}{3\left(x-2\right)}\)

\(\dfrac{x^2-9y^2}{x^2+xy-6y^2}=\dfrac{\left(x-3y\right)\left(x+3y\right)}{\left(x-2y\right)\left(x+3y\right)}=\dfrac{x-3y}{x-2y}\)

a) ĐKXĐ: 

\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)

b) \(A=\dfrac{x^2-2x+1}{x^2-1}\)

\(A=\dfrac{x^2-2\cdot x\cdot1+1^2}{x^2-1^2}\)

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

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

c) Thay x = 3 vào A ta có:

\(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)

a) ĐKXĐ: 

\(9x^2-y^2\ne0\Leftrightarrow\left(3x\right)^2-y^2\ne0\Leftrightarrow\left(3x-y\right)\left(3x+y\right)\ne0\)

\(\Leftrightarrow3x\ne\pm y\) 

b) \(B=\dfrac{6x-2y}{9x^2-y^2}\)

\(B=\dfrac{2\cdot3x-2y}{\left(3x\right)^2-y^2}\)

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

\(B=\dfrac{2}{3x+y}\)

Thay x = 1 và \(y=\dfrac{1}{2}\) và B ta có:

\(B=\dfrac{2}{3\cdot1+\dfrac{1}{2}}=\dfrac{2}{3+\dfrac{1}{2}}=\dfrac{2}{\dfrac{7}{2}}=\dfrac{4}{7}\)

tách ra <=> \(\frac{X^2-1+x-1}{X^2+2x+5x+10}\) <=> \(\frac{\left(x^2-1\right)+\left(x-1\right)}{\left(x^2+2x\right)+\left(5x+10\right)}< =>\frac{\left(x-1\right)\left(x+1\right)+\left(x-1\right)}{x\left(x+2\right)+5\left(x+2\right)}\)

Đặt nhân tử chung đi rồi tính nốt:v

\(\frac{x^2+x-2}{x^2+7x+10}=\frac{x^2+2x-x-2}{x^2+5x+2x+10}=\frac{x\left(x+2\right)-\left(x+2\right)}{x\left(x+5\right)+2\left(x+5\right)}=\frac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}=\frac{x-1}{x+5}\)

\(M=\dfrac{x^2}{x^2-3x}\left(x\ne0;x\ne3\right)\\ M=\dfrac{x^2}{x\left(x-3\right)}\\ M=\dfrac{x}{x-3}\)

\(N=\dfrac{x}{x+1}+\dfrac{3x+1}{x^2-1}\left(x\ne\pm1\right)\\ N=\dfrac{x-1+3x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x}{\left(x+1\right)\left(x-1\right)}\)

a) M=\(\dfrac{x^2}{x^2-3x}\)=\(\dfrac{x.x}{x\left(x-3\right)}\)=\(\dfrac{x}{x-3}\)

b)\(\dfrac{x}{x+1}+\dfrac{3x+1}{x^2-1}\)=\(\dfrac{2x^2+4x+1}{x^3+x^2}\)

\(M=\dfrac{x^2}{x^2-3x}=\dfrac{x}{x-3}\)

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

\(1,\Leftrightarrow x=10\\ 2,=x^2-4x+4-x^2+4x=4\\ 3,=\left(x+y\right)^2-49=\left(x+y+7\right)\left(x+y-7\right)\)

Câu 1 :

\(2x-20=0\)

\(2x=0+20\)

\(2x=20\)

\(2.x=20\)

\(x=20:2\)

\(x=10\)

\(a,=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}=\dfrac{x+1}{x}\\ b,=\dfrac{-\left(x^2-5x-6\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+1\right)\left(x-6\right)}{\left(x+2\right)^2}\)