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

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

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

\(=x^6-64\)

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

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

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

\(=x^6-64\)

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

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

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

\(=x^6+64\)

phần dưới là tìm x

\(\dfrac{5x+2}{x^2-4}+\dfrac{x-5}{x-2}=\dfrac{5x+2+x^2-3x-10}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+4}{x+2}\\ \left(x+4\right)^2-\left(x+3\right)\left(x-2\right)=-13\\ \Leftrightarrow x^2+8x+16-x^2+x+6=-13\\ \Leftrightarrow9x=-13-22=-35\\ \Leftrightarrow x=-\dfrac{35}{9}\)

Bài 2:

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

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

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

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

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

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

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

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

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)

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

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

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

a: \(\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)

\(=\dfrac{4-x^2-2x+2x^2+5-4x}{x-3}=\dfrac{x^2-6x+9}{x-3}\)

=(x-3)^2/(x-3)

=x-3

b: \(\dfrac{2}{x+2}+\dfrac{-4}{2-x}+\dfrac{5x+2}{4-x^2}\)

\(=\dfrac{2}{x+2}-\dfrac{4}{x-2}-\dfrac{5x+2}{x^2-4}\)

\(=\dfrac{2x-4-4x-8-5x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-7x-14}{\left(x-2\right)\left(x+2\right)}\)

=-7(x+2)/(x-2)(x+2)

=-7/(x-2)

a) Ta có: \(\left(2x+3\right)^2-4\left(x-2\right)\left(x+2\right)\)

\(=4x^2+12x+9-4\left(x^2-4\right)\)

\(=4x^2+12x+9-4x^2+16\)

\(=12x+25\)

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

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

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

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

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

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

\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+5\right)\)

\(=x^3-8-x^3-5\)

=-13

\(\left(x-2\right)\cdot\left(x^2+2x+4\right)-\left(x^3+5\right)\\ =x^2-8-x^3-5\\ =-13\)

Vậy \(\left(2x^4+2x^3+3x^2-5x-20\right):\left(x^2+x+4\right)=2x^2-5\)

Bài 1: 

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

Bài 2: 

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

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)