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

Vậy giá trị nhỏ nhất của biểu thức trên là -1 khi x=-1.

c: \(=\left(x+1\right)^2+1>0\forall x\)

Trả lời:

a, \(x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTNN của biểu thức bằng 2 khi x = 3

b, \(-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x^2-6x+9+2\right)=-\left[\left(x-3\right)^2+2\right]\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTLN của biểu thức bằng - 2 khi x = 3

c, \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\forall x\inℤ\)  (đpcm)

Dấu "=" xảy ra khi x + 1 = 0 <=> x = - 1

Áp dụng Bunyakovsky, ta có :

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

=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

Ta có:

\(M=x^2-2x\left(y+1\right)+3y^2+2025\)

\(M=x^2-2\cdot x\cdot\left(y+1\right)+\left(y+1\right)^2+3y^2+2025-\left(y+1\right)^2\) 

\(M=\left[x-\left(y+1\right)\right]^2+3y^2+2025-y^2-2y-1\)

\(M=\left(x-y-1\right)^2+2y^2-2y+2024\)

\(M=\left(x-y-1\right)^2+2\left(y-\dfrac{1}{2}\right)^2+\dfrac{4047}{2}\)

Mà: \(\left\{{}\begin{matrix}\left(x-y-1\right)^2\ge0\\2\left(y-\dfrac{1}{2}\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow M=\left(x-y-1\right)^2+2\left(y-\dfrac{1}{2}\right)^2+\dfrac{4047}{2}\ge\dfrac{4047}{2}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x-y-1=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}+1\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\) 

Vậy GTNN của M là .... 

Đặt \(A=\frac{x^2-2x+2020}{x^2}\)

\(=1-\frac{2}{x}+\frac{2020}{x^2}\)

\(=2020\left(\frac{1}{x^2}-\frac{1}{1010\cdot x}+\frac{1}{2020}\right)\)

\(=2020\left(\frac{1}{x^2}-2\cdot\frac{1}{x}\cdot\frac{1}{2020}+\frac{1}{2020^2}+\frac{1}{2020}-\frac{1}{2020^2}\right)=2020\left(\frac{1}{x}-\frac{1}{2020}\right)^2+1-\frac{1}{2020}=2020\left(\frac{1}{x}-\frac{1}{2020}\right)^2+\frac{2019}{2020}\ge\frac{2019}{2020}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\frac{1}{x}-\frac{1}{2020}=0\)

=>x=2020

ĐKXĐ: x<>0

Ta có: \(A=\frac{x^2-2x+7}{x^2}\)

\(=1-\frac{2}{x}+\frac{7}{x^2}\)

\(=7\left(\frac{1}{x^2}-\frac27\cdot\frac{1}{x}+\frac17\right)\)

\(=7\left(\frac{1}{x^2}-2\cdot\frac{1}{x}\cdot\frac17+\frac{1}{49}+\frac{6}{49}\right)=7\left(\frac{1}{x}-\frac17\right)^2+\frac67\ge\frac67\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\frac{1}{x}-\frac17=0\)

=>\(\frac{1}{x}=\frac17\)

=>x=7(nhận)

ĐKXĐ: x<>0

Ta có: \(A=\frac{x^2-2x+7}{x^2}\)

\(=1-\frac{2}{x}+\frac{7}{x^2}\)

\(=7\left(\frac{1}{x^2}-\frac27\cdot\frac{1}{x}+\frac17\right)\)

\(=7\left(\frac{1}{x^2}-2\cdot\frac{1}{x}\cdot\frac17+\frac{1}{49}+\frac{6}{49}\right)=7\left(\frac{1}{x}-\frac17\right)^2+\frac67\ge\frac67\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\frac{1}{x}-\frac17=0\)

=>\(\frac{1}{x}=\frac17\)

=>x=7(nhận)

\(a,-x^2+2x+5=-\left(x^2-2x-5\right)=-\left(x^2-2x+1-6\right)=-\left(x-1\right)^2+6\le6\)

dấu'=' xảy ra<=>x=1=>Max A=6

\(b,B=-x^2-y^2+4x+4y+2=-x^2+4x-4-y^2+4x-4+10\)

\(=-\left(x^2-4x+4\right)-\left(y^2-4x+4\right)+10\)

\(=-\left(x-2\right)^2-\left(y-2\right)^2+10=-\left[\left(x-2\right)^2+\left(y-2\right)^2\right]+10\le10\)

dấu"=" xảy ra<=>x=y=2=>Max B=10

\(c,C=x^2+y^2-2x+6y+12=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

dấu'=' xảy ra<=>x=1,y=-3=>MinC=2