a) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)

\(=x\left(x^2-5x+1\right)-2\left(x^2-5x+1\right)-x\left(x^2+11\right)\)

\(=x^3-5x^2+x-2x^2+10x-2-x^3-11x\)

\(=-7x^2-2\)

b) \(\left(x-1\right)\left(x^2+x+1\right)+x^3-2\)

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

\(=x^3+x^2+x-x^2-x-1+x^3-2\)

\(=2x^3-3\)

c) \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)

\(=x\left(x+y\right)-y\left(x+y\right)-2x\left(x-y\right)\)

\(=x^2+xy-yx-y^2-2x^2+2xy\)

\(=-x^2-y^2+2xy\)

a, \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)

\(=x^3-7x^2+11x-2-x^3-11x=-7x^2-2\)

b, \(\left(x-1\right)\left(x^2+x+1\right)+\left(x^3-2\right)\)

\(=x^3-1+x^3-2=2x^3-3\)

c, \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)

\(=x^2-y^2-2x^2+2xy=-x^2-y^2+2xy\)

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

\(\dfrac{x^2-4}{3x+6}=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x-2}{3}\)

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

b. \(\dfrac{x^2-4}{3x+6}=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x-2}{3}\)

\(A=\dfrac{x-9}{3+\sqrt{x}}\) (đề như này pk?)

a) Để A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\3+\sqrt{x}\ne0\left(lđ\right)\end{matrix}\right.\)\(\Rightarrow x\ge0\)

b) \(A=\dfrac{x-9}{3+\sqrt{x}}=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{3+\sqrt{x}}=\sqrt{x}-3\)

c) Với x=0 (tmđk) thay vào A ta được: \(A=\sqrt{0}-3=-3\)

Với x=-1 (ktm đk)

Với x=16 (tmđk) thay vào A ta được: \(A=\sqrt{16}-3=1\)

d) \(A\in Z\Leftrightarrow\sqrt{x}-3\in Z\Leftrightarrow\sqrt{x}\in Z\) \(\Leftrightarrow\) x là số chính phương

\(a,=x^2-3x-10-x^2+3x=-10\\ b,=\left(x+1\right)\left(x+1-x+1\right)=2\left(x+1\right)=2x+2\)

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

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

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

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

b: |x-2|=2

=>x-2=2 hoặc x-2=-2

=>x=0(nhận) hoặc x=4(nhận)

Khi x=0 thì \(A=\dfrac{2\cdot0}{0-3}=\dfrac{-2}{3}\)

Khi x=4 thì \(A=\dfrac{2\cdot4}{4-3}=8\)

c: A>0

=>x/x-3>0

=>x>3 hoặc x<0

=>x>3

a) \(A=\left(x-1\right).\left(x+1\right)+\left(x+2\right).\left(x^2+2x+4\right)-x.\left(x^2+x+2\right)\)

\(=x^2-1+x^3+2x^2+4x+2x^2+4x+8-x^3-x^2-2x\)

\(=\left(x^3-x^3\right)+\left(x^2+2x^2+2x^2-x^2\right)+\left(4x+4x-2x\right)+\left(-1+8\right)\)

\(=4x^2+6x+7\)

b) Thay vào ta được

\(A=4.\left(\frac{1}{2}\right)^2+6.\frac{1}{2}+7=1+3+7=11\)

bài 1 : a. x^3 +27 -54-x^3 =-27

b. 8x^3 +y^3 -8x^3 +y^3 =2y^3

c. (2x-1+2x+2)(2x-1-2x-2)=(4x+1).(-3)=-12x-3

d. a^3 +b^3 +3ab(a+b) -3ab(a+b)=a^3+b^3

 a. (x-1)^2 =5^2

x-1=5

x=6

A=(4x^2 +4x+1 )+1 

A=(2x+1)^2 +1 >0

B=(x^2 -2x+1 )+x^2

B=(x-1)^2 +x^2 >0