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 

\(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(=\sqrt{\left(x-1\right)^2}+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-4\right|+\left(\left|x-1\right|+\left|x-6\right|\right)\)

\(=\left|x-4\right|+\left(\left|x-1\right|+\left|6-x\right|\right)\)

Ta có \(\hept{\begin{cases}\left|x-4\right|\ge0\forall x\\\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=\left|5\right|=5\end{cases}}\)

=> \(\left|x-4\right|+\left(\left|x-1\right|+\left|6-x\right|\right)\ge5\forall x\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-4=0\\\left(x-1\right)\left(6-x\right)\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\1\le x\le6\end{cases}}\Leftrightarrow x=4\)

=> MinA = 5 <=> x = 4

Ta có: \(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(\Rightarrow A=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-4\right|+\left|x-1\right|+\left|x-6\right|\)

Xét \(\left|x-1\right|+\left|x-6\right|\)ta có: 

\(\left|x-1\right|+\left|x-6\right|=\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=\left|5\right|=5\)(1)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(6-x\right)\ge0\)

TH1: Nếu \(\hept{\begin{cases}x-1< 0\\6-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\6< x\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x>6\end{cases}}\)( vô lý )

TH2: Nếu \(\hept{\begin{cases}x-1\ge0\\6-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\6\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le6\end{cases}}\Leftrightarrow1\le x\le6\)

mà \(\left|x-4\right|\ge0\)(2)

Từ (1) và (2) \(\Rightarrow A\ge5\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-4=0\\1\le x\le6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\1\le x\le6\end{cases}}\Leftrightarrow x=4\)

Vậy \(minA=5\)\(\Leftrightarrow x=4\)

Xét biểu thức \(A=x\left(x-3\right)\left(x-4\right)\left(x-7\right)=\left(x^2-7x\right)\left(x^2-7x+12\right)\)

Đặt \(x^2-7x+6\rightarrow t\)Khi đó \(A=\left(t-6\right)\left(t+6\right)=t^2-36\ge-36\)

Dấu "=" xảy ra khi và chỉ khi \(t=0\)hay \(x^2-7x+6=0=>\left(x-6\right)\left(x-1\right)=0=>\orbr{\begin{cases}x=6\\x=1\end{cases}}\)

Vậy GTNN của biểu thức \(A=-36\)đạt được khi \(x=6orx=1\)

Xét biểu thức \(B=2x^2+y^2-2xy-2x+3=\left(x^2-2xy+y^2\right)+x^2-2x+1+2\)

\(=\left(x-y\right)^2+\left(x-1\right)^2+2\ge2\)

Dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}x-y=0\\x-1=0\end{cases}< =>\hept{\begin{cases}1-y=0\\x=1\end{cases}}< =>\hept{\begin{cases}x=1\\y=1\end{cases}< =>x=y=1}}\)

Vậy GTNN của biểu thức \(B=2\)đạt được khi \(x=y=1\)

Xét biểu thức \(C=x^2+y^2-3x+3y=\left(x^2-3x+\frac{9}{4}\right)+\left(y^2+3y+\frac{9}{4}\right)-\frac{9}{2}\)

\(=\left(x^2-3x+\frac{3^2}{2^2}\right)+\left(y^2+3y+\frac{3^2}{2^2}\right)-\frac{9}{2}=\left(x-\frac{3}{2}\right)^2+\left(y+\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)

Dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}x-\frac{3}{2}=0\\y+\frac{3}{2}=0\end{cases}}< =>\hept{\begin{cases}x=\frac{3}{2}\\y=-\frac{3}{2}\end{cases}< =>x=-y=\frac{3}{2}}\)

Vậy GTNN của biểu thức \(C=-\frac{9}{2}\)đạt được khi \(x=-y=\frac{3}{2}\)

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

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 .... 

\(a,=x^2+2x+1+2019=\left(x+1\right)^2+2019\ge2019\) dấu"=" xảy ra<=>x=-1

b,\(=m^2+2.2m+4-5=\left(m+2\right)^2-5\ge-5\) dấu"=" xảy ra<=>m=-2

c, \(=x-2\sqrt{x}+10=x-2\sqrt{x}+1+9=\left(\sqrt{x}-1\right)^2+9\ge9\)

dấu"=" xảy ra<=>x=1

b, \(4x-8\sqrt{x}+2020=4x-2.2.2\sqrt{x}+4+2016=\left(2\sqrt{x}-2\right)^2+2016\ge2016\)

dấu"=" xảy ra<=>x=1

a) Ta có: \(x^2+2x+2020\)

\(=x^2+2x+1+2019\)

\(=\left(x+1\right)^2+2019\ge2019\forall x\)

Dấu '=' xảy ra khi x=-1

b) Ta có: \(m^2+4m-1\)

\(=m^2+4m+4-5\)

\(=\left(m+2\right)^2-5\ge-5\forall m\)

Dấu '=' xảy ra khi m=-2

c) Ta có: \(m^2+m\)

\(=m^2+2\cdot m\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}\)

\(=\left(m+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\forall m\)

Dấu '=' xảy ra khi \(m=-\dfrac{1}{2}\)

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

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

b) Ta có: \(B=x^2+2x+y^2-4y+6\)

\(=x^2+2x+1+y^2-4y+4+1\)

\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)

c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)

\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)

\(B=x^2+2x+y^2-4y+6\)

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

\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2

\(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)

dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)

\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)

=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)

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

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

dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)