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

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

\(\Rightarrow-A=\left(x^2-2x+1\right)+\left(4y^2+4y+1\right)-7\)

\(\Rightarrow-A=\left(x-1\right)^2+\left(2y+1\right)^2-7\)

Vay \(A_{max}=7\Leftrightarrow x=1;y=-\frac{1}{2}\)

\(A=x-x^2\)

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

\(-A=x^2-2\cdot\frac{1}{2}\cdot x+\frac{1}{4}-\frac{1}{4}\)

\(-A=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\)

\(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(\Rightarrow-A\ge-\frac{1}{4}\)

\(\Rightarrow A\le\frac{1}{4}\)

dấu "=" xảy ra khi : 

\(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

\(x^2+2xy-8y^2+2xz+14yz-3z^2\)

\(=\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(-9x^2+12yz-4x^2\right)\)

\(=\left(x+y+z\right)^2-\left[\left(3x\right)^2-2.3x.2y+\left(2x\right)^2\right]\).

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

\(=\left(x+y+z-3y+2x\right)\left(x+y+z+3y-2x\right)\)

\(A=\left(1+b^2+a^2+a^2b^2\right).\left(1+c^2\right)\)

\(=1+a^2+b^2+c^2+a^2c^2+b^2c^2+a^2b^2+a^2b^2c^2\)

\(=1+\left(a+b+c\right)^2-2.\left(ab+bc+ac\right)+\left(ab+bc+ac\right)^2-2abc.\left(a+b+c\right)+a^2b^2c^2\)

Thay ab+bc+ac=1 vào A, ta có:

\(A=1+\left(a+b+c\right)^2-2+1-2abc.\left(a+b+c\right)+a^2b^2c^2\)

\(=\left(a+b+c\right)^2-2abc.\left(a+b+c\right)+a^2b^2c^2\)

\(=\left(a+b+c-abc\right)^2\)

Vì a,b,c thuộc Z 

\(\Rightarrow\left(a+b+c-abc\right)^2\)là số chính phương

\(\hept{\begin{cases}\left(1+a^2\right)=\left(ab+bc+ca+a^2\right)=b\left(a+c\right)+a\left(a+c\right)=\left(a+b\right)\left(a+c\right)\\\left(1+b^2\right)=\left(ab+bc+ca+b^2\right)=a\left(b+c\right)+b\left(b+c\right)=\left(a+b\right)\left(b+c\right)\\\left(1+c^2\right)=\left(ab+bc+ca+c^2\right)=a\left(b+c\right)+c\left(b+c\right)=\left(a+c\right)\left(b+c\right)\end{cases}}\)

\(\Rightarrow A=\text{[}\left(a+b\right)\left(b+c\right)\left(c+a\right)\text{]}^2\Rightarrow\text{đ}pcm\)

\(a,x^3-3x^2-x-3\)

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

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

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

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

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

\(b,A=3+2x-x^2\)

\(-A=x^2-2x-3\)

\(-A=x^2-2\cdot x+1-4\)

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

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

\(-A=\left(x-3\right)\left(x+1\right)\)

\(A=-\left(x-3\right)\left(x+1\right)\)

\(P=\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(P=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x\)

\(P=\left(x^3+x^3-2x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x-24x\right)+\left(8-8\right)\)

\(P=0+0+0+0\)

\(P=0\)

=> P không phụ thuộc vào biến