\(A=\left(6x-2\right)^2+2\left(6x-2\right)\left(2-5x\right)+\left(2-5x\right)^2\)

\(=\left(6x-2+2-5x\right)^2=x^2\)

\(B=\left(2a^2+1+2a\right)\left(2a^2+1-2a\right)-\left(2a^2+1\right)^2\)

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

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

1.\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3x+9}\right)=\left(\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\right):\left(\frac{3x-9-x^2}{3x\left(x+3\right)}\right)=-\frac{1}{x-3}\)

2.\(\left(\frac{2\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}=\frac{4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}=\frac{x-2}{2}\)

3.\(\left(\frac{3\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(3x+1\right)}\right):\frac{2x\left(x+5\right)}{\left(1-3x\right)^2}=\frac{-6x^2+11x+3}{\left(1-3x\right)\left(3x+1\right)}.\frac{\left(1-3x\right)^2}{2x\left(x+5\right)}=\frac{-6x^2+11x+3}{\left(3x+1\right)}.\frac{\left(1-3x\right)}{2x\left(x+5\right)}\)

4.\(\left(\frac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}+\frac{x}{5-x}=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}=\frac{5}{x-5}-\frac{x}{x-5}=-1\)

\(1.A=14x^2+16x-21x-24=14x^2-5x-24\)

\(2.A=3x^3-2x^2-15x+14\)

\(3.A=x^4-x^3+2x^2+7x-5\)

\(4.A=x^4-7x^3+3x^2+21x-18\)

\(5.A=4x^2-y^2\)

\(6.A=x^3-8y^3\)

\(7.A=x^3+27y^3\)

\(8.A=x^2y^2-1\)

3) \(=\frac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\frac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)\(=\frac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}.\frac{\left(1-3x\right)^2}{2x\left(3x+5\right)}=\frac{1-3x}{2\left(1+3x\right)}\)

4) \(=\frac{x.x-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}.\frac{x^2+5x}{2x-5}+\frac{x}{5-x}\)\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)

\(=\frac{5}{x-5}+\frac{-x}{x-5}=\frac{5-x}{x-5}=-1\)

5) \(=\left[\frac{x^2+xy}{\left(x+y\right)\left(x^2+y^2\right)}+\frac{y}{x^2+y^2}\right]:\left[\frac{1}{x-y}-\frac{2xy}{\left(x-y\right)\left(x^2+y^2\right)}\right]\)

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

câu 3 đến câu 5 là được rồi ạ

Bài 1 a 

\(x^4+3x^3-6x^2-8x=x\left(x^3-8\right)+3x^2\left(x-2\right)\)

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

\(=\left(x-2\right)\left[x\left(x^2+2x+4\right)+3x^2\right]\)

\(=x\left(x-2\right)\left(x^2+5x+4\right)=x\left(x-2\right)\left(x+1\right)\left(x+4\right)\)

Bài 2 : 

a, \(8x^3y^5z:2x^2y^3=4xy^2z\)

b, \(\frac{1}{2}x^3y^5:\left(-\frac{3}{4x^2y^3}\right)=-\frac{2x^5y^8}{3}\)

c, \(\left(5x^5-x^3+3x^2\right):3x^2=\frac{5}{3}x^3-3x+1\)

\(B=x^2-x+13\)là số chính phương \(\Leftrightarrow4B=4x^2-4x+52\)là số chính phương.

\(4x^2-4x+52=n^2\)

\(\Leftrightarrow\left(2x-1\right)^2+51=n^2\)

\(\Leftrightarrow\left(n-2x+1\right)\left(n+2x-1\right)=51=1.51=3.17\)

Ta có bảng giá trị: 

Vậy \(x\in\left\{-12,-3,4,13\right\}\)thỏa mãn ycbt.