\(\left(x+2\right)^3-x.\left(x+2\right).\left(x-2\right)+6x^2\)

\(=x^3+3x^2.2+3x.2^2+2^3-x.\left(x^2-2^2\right)+6x^2\)

\(=x^3+6x^2+12x+8-\left(x^2-4\right)+6x^2\)

\(=x^3+6x^2+12x+8-x^3+4x+6x^2\)

\(=\left(x^3-x^3\right)+\left(6x^2+6x^2\right)+\left(12x+4x\right)+8\)

\(=12x^2+16x+8\)

\(x^2-\left(5-y\right)^2\)

\(=[x+\left(5-y\right)].[x-\left(5-y\right)]\)

\(=\left(x+5-y\right).\left(x-5+y\right)\)

\(=\left(x-y+5\right).\left(x+y-5\right)\)

t hong bíc nè 

\(A=x^2+y^2+z^2-yz-4x-3y+2027\)

\(\Rightarrow4A=4x^2+4y^2+4z^2-4yz-16x-12y+8108\)

\(=\left(4x^2-16x+16\right)+\left(3y^2-12y+12\right)+\left(y^2-4yz+4z^2\right)+8080\)

\(=4.\left(x^2-4x+4\right)+3.\left(y^2-4y+4\right)+\left(y-2z\right)^2+8080\)

\(=4.\left(x-2\right)^2+3.\left(y-2\right)^2+\left(y-2z\right)^2+8080\)

Mà: \(\hept{\begin{cases}4.\left(x-2\right)^2\ge0\\3.\left(y-2\right)^2\ge0\\\left(y-2z\right)^2\ge0\end{cases}}\)

\(\Rightarrow4.\left(x-2\right)^2+3.\left(y-2\right)^2+\left(y-2z\right)^2\ge0\)

\(\Rightarrow4.\left(x-2\right)^2+3.\left(y-2\right)^2+\left(y-2z\right)^2+8080\ge8080\)

\(\Rightarrow A\ge8080\)

Dấu '' = '' xảy ra khi: 

\(\hept{\begin{cases}4.\left(x-2\right)^2=0\\3.\left(y-2\right)^2=0\\\left(y-2z\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=2\\z=1\end{cases}}\)

Vậy giá trị nhỏ nhất của \(A=2020\) khi \(\hept{\begin{cases}x=y=2\\z=1\end{cases}}\)

\(A=a^4+6a^3+11a^2+6a=a\left(a^3+a^2+5a^2+5a+6a+6\right)\)

\(=a\left(a+1\right)\left(a^2+5a+6\right)=a\left(a+1\right)\left(a+2\right)\left(a+3\right)\)

suy ra \(A\)là tích của \(4\)số nguyên liên tiếp, 

Do đó trong \(a,a+1,a+2,a+3\)có ít nhất \(1\)thừa số chia hết cho \(3\). \(1\)thừa số chia hết cho \(4\)và \(1\)thừa số chia hết cho \(2\)nhưng không chia hết cho \(4\).

Khi đó \(A\)chia hết cho \(3.4.2=24\).

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

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

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

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

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

\(x^5+x+1\)

\(=x^5-x^2+x+1\)

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

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

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

\(x^3+5x^2+5x+1\)

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

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

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

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

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

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

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

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

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