a: Ta có: \(\left(x+1\right)^2+\left(x-1\right)^2-2\left(1+x\right)\left(1-x\right)\)

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

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

\(=4x^2\)

c: Ta có: \(3\left(x+2\right)^2-\left(3x+1\right)\left(x+5\right)+\left(x+5\right)^2\)

\(=3x^2+12x+12-3x^2-16x-5+x^2+10x+25\)

\(=x^2+6x+32\)

a: Đặt f(x)=0

=>\(-3x^2+2x=0\)

=>\(3x^2-2x=0\)

=>x(3x-2)=0

=>\(\left[{}\begin{matrix}x=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Bảng xét dấu:

b: Đặt G(x)=0

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5

Bảng xét dấu:

c: Đặt H(x)=0

=>\(4x^2-4x+1=0\)

=>\(\left(2x-1\right)^2=0\)

=>2x-1=0

=>x=1/2

Bảng xét dấu:

d: Đặt Q(x)=0

=>(2x+3)(x-5)=0

=>\(\left[{}\begin{matrix}2x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=5\end{matrix}\right.\)

Bảng xét dấu:

a: Ta có: \(\left(x-2\right)^2-\left(2x-1\right)^2+\left(3x-1\right)\left(x-5\right)\)

\(=x^2-4x+4-4x^2+4x-1+3x^2-15x-x+5\)

\(=-16x+8\)

b: Ta có: \(\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

=27x-55

a) (x+2)²+x(x-4)

   =x²+4x+4+x²-4x

   =2x²+4

b)(x-3)²-(x+3)(x-4)

  =x²-6x+9-x²+4x-3x+12

  =-5x+12

c) (3x+1)²+3x(2-4x)

   =9x²+6x+1+6x-12x²

   =-3x²+12x+1

d) (2x-4y)²-(2x-3)(2x-3y)

  =4x²-16xy+16y²-4x²+6xy+6x-9y

  =16y²-10xy+6x-9y

\(A=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)ĐK : \(x\ne-2;2\)

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

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

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

\(=\dfrac{x^3-4x+x^2-2x+24}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{x^3+x^2-6x+24}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}\)

\(=\dfrac{x^3+x^2-6x+24}{6\left(x-4\right)\left(x-2\right)}=\dfrac{\left(x+4\right)\left(x^2-3x+6\right)}{6\left(x-4\right)\left(x-2\right)}\)

P/s : mình thấy đề này cứ sai sai ở đâu ý ! 

b, Ta có : \(\dfrac{\left(x+4\right)\left(x^2-3x+6\right)}{6\left(x-4\right)\left(x-2\right)}=2\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x^2-3x+6\right)-12\left(x-4\right)\left(x-2\right)}{6\left(x-4\right)\left(x-2\right)}=0\)

\(\Rightarrow x^3-11x^2+66x-72=0\)

a: Ta có: \(\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)

\(=\left(x^4-4\right)\left[\left(x^2+2\right)^2-4x^2\right]\)

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

\(=\left(x^4-4\right)\cdot\left(x^4+4\right)\)

\(=x^8-16\)

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

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

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

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

\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)

ĐKXĐ: \(x\notin\left\{-1;2;-2\right\}\)

a) Ta có: \(A=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)

\(=\left(\dfrac{\left(x+1\right)^2}{x^2-x+1}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)

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

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

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

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

b) Để A nguyên thì \(3x⋮x-2\)

\(\Leftrightarrow3x-6+6⋮x-2\)

mà \(3x-6⋮x-2\)

nên \(6⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(6\right)\)

\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)

Kết hợp ĐKXĐ, ta được:

\(x\in\left\{3;1;4;0;5;8;-4\right\}\)

Vậy: Để A nguyên thì \(x\in\left\{3;1;4;0;5;8;-4\right\}\)