1/ n=3

\(B=x^2+\frac{1}{x^2}\ge\sqrt{x^2\cdot\frac{1}{x^2}}=1\)

Dấu "=" xảy ra tại \(x=y=1\)

\(\left(A+B\right)^2=A^2+2AB+B^2\)

\(\left(A+B\right)^3=A^3+3A^2B+3AB^2+B^3\)

\(A^2-B^2=\left(A-B\right)\left(A+B\right)\)

2

a

\(15x^2y^3z^2-20x^2yz^2+10xy^3z\)

\(=5xyz\left(3xy^2z-4xz+2y^2\right)⋮5xyz\)

b

\(13ab^2+abc+32a=a\left(13b^2+bc+32\right)\) 

TH1:\(13b^2+bc+32=7b\cdot P\left(x\right)\) thì A chia hết cho B

TH2:\(13b^2+bc+32=7b\cdot Q\left(x\right)+r\left(r>0\right)\) thì A không chia hết cho B

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

(Dấu "="\(\Leftrightarrow x=0\))

Vậy \(A_{min}=-4\Leftrightarrow x=0\)

\(A=\left(x+1\right)\left(x-3\right)\left(x+2\right)\left(x+6\right)\)

\(=\left(x+1\right)\left(x+2\right)\left(x+6\right)\left(x-3\right)\)

\(=\left(x^2+3x+2\right)\left(x^2+3x-18\right)\)

Đặt \(x^2+3x+2=t\)

\(\Rightarrow BT=t\left(t-20\right)=\left(t-10\right)^2-100\ge-100\)

a)\(A=5-8x-2x^2\)

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

\(=-2\left(x^2+4x+4-\frac{13}{2}\right)\)

\(=-2\left[\left(x+2\right)^2-\frac{13}{2}\right]\)

\(=-2\left[\left(x+2\right)^2\right]+13\le13\)

Vậy \(A_{max}=13\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

a) \(A=5x^2-4x+1\)

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

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

\(=5\left[\left(x-\frac{2}{5}\right)^2-\frac{2}{25}\right]\)

\(=5\left[\left(x-\frac{2}{5}\right)^2\right]-2\ge-2\)

Vậy \(A_{min}=-2\Leftrightarrow x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}\)

Sửa)):Dòng 3

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

\(=5\left[\left(x-\frac{2}{5}\right)^2+\frac{1}{25}\right]\)

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

(Dấu "="\(\Leftrightarrow x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}\)

a) \(x^2-10x+4y^2-4y+26=0\)

\(\Leftrightarrow\left(x^2-10x+25\right)+\left(4y^2-4y+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)^2+\left(2y-1\right)^2=0\)

Mà \(\Leftrightarrow\left(x-5\right)^2+\left(2y-1\right)^2\ge0\)

Dấu "="\(\Leftrightarrow\hept{\begin{cases}x-5=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{1}{2}\end{cases}}\)

Gọi A là biểu thức trên

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

Với \(x=\frac{2}{3},y=1-3=-2\Rightarrow A=\frac{8}{27}+8=\frac{210}{27}\)