Câu 1: C

Câu 2: =x(x-2)*(x+2)

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 : 

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

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

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

b: Khi x=3 thì \(A=\dfrac{-3}{3^2+1}=-\dfrac{3}{10}\)

c: x^2+1>=0

=>3/x^2+1>=0

=>-3/x^2+1<=0

=>A<=0(ĐPCM)

B1:

\(a,A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(=\left(\frac{\left(3-x\right)\left(x+3\right)^2}{\left(x+3\right)\left(x^2-9\right)}+\frac{x}{x+3}\right).\frac{x+3}{3x^2}\)

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

\(=\left(\frac{\left(3-x\right)\left(x+3\right)}{x^2-9}+\frac{x\left(x-3\right)}{x^2-9}\right).\frac{x+3}{3x^2}\)

\(=\frac{3x+9-x^2-3x+x^2-3x}{x^2-9}.\frac{x+3}{3x^2}\)

\(=\frac{9-3x}{x^2-9}.\frac{x+3}{3x^2}\)

\(=\frac{3\left(3-x\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)3x^2}\)

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

B2: 

\(a,B=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right)\)

\(=\left(\frac{x}{x^2-4}-\frac{2\left(x+2\right)}{x^2-4}+\frac{x+2}{x^2-4}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)

\(=\left(\frac{x-2x-4+x-2}{x^2-4}\right):\frac{6}{x+2}\)

\(=-\frac{6}{x^2-4}.\frac{x+2}{6}\)

\(=\frac{-6\left(x+2\right)}{\left(x+2\right)\left(x-2\right)6}=-\frac{1}{x-2}\)

cn ĐKXĐ và phần b,c của cả 2 bài,bn tự lm nốt

\(P=\dfrac{3x^2+6x+3}{x+1}\)

\(a,\) Điều kiện xác định: \(x+1\ne0\Leftrightarrow x\ne-1\)

\(b,P=\dfrac{3x^2+6x+3}{x+1}=\dfrac{3\left(x^2+2x+1\right)}{x+1}=\dfrac{3\left(x+1\right)^2}{x+1}=3\left(x+1\right)=3x+3\)

\(c,x=1\Rightarrow P=3.1+3=6\)

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

b: Khi x=1 thì P=3+3=6

c: P<0

=>x+1<0

=>x<-1

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

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

c: Thay x=2 vào A, ta được:

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

d: Để A=2 thì x+1=2x-2

=>-x=-3

hay x=3(nhận)

a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)

Bạn có thể giúp mình 2 câu còn lại dc kh ạ 

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

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

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

`a, (3x^2+6xy)/(6x^2) = (x+2y)/(3x)`

`b, (2x^2-x^3)/(x^2-4) = (x^2(2-x))/((x-2)(x+2))`

`= -x^2/(x+2)`

`c, (x+1)/(x^3+1) = 1/(x^2-x+1)`