\(a,x^2+y^2-4x-2y+6\)

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

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

Ta có: \(\left(x-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

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

Hay: \(x^2+y^2-4x-2y+6\ge1\)

\(b,x^2+4y^2+z^2-4x+4y-8z+25\)

\(=\left(x^2-4x+4\right)+\left(4y^2+4y+1\right)+\left(z^2-8z+16\right)+4\)

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

Vì: \(\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2\ge0\forall x,y,z\)

\(\Rightarrow\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2+4\ge4\forall x,y,z\)

Hay: \(x^2+4y^2+z^2-4x+4y-8z+25\ge4\)

=.= hok tốt !!

Chúc bạn có 1 ngày vui vẻ!!!

a) \(A=x^2+2x+2\)

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

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

b) \(B=4x^2-4x+11\)

\(=4x^2-4x+1+10\)

\(=\left(2x-1\right)^2+10>0\forall x\)

c) \(C=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

d) Ta có: \(D=x^2-2x+y^2+4y+6\)

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

\(=\left(x-1\right)^2+\left(y+2\right)^2+1>0\forall x,y\)

e) Ta có: \(D=x^2-2xy+y^2+x^2-8x+20\)

\(=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4>0\forall x,y\)

\(A=x^2+2x+2=\left(x+1\right)^2+1\ge1>0\left(\forall x\right)\)

\(B=4x^2-4x+11=\left(2x-1\right)^2+10\ge10>0\left(\forall x\right)\)

\(C=x^2-x+1=x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1>0\)

\(E=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4\ge4>0\)\(\left(\forall x,y\right)\)

Áp dụng BĐT Cô-si với 2 số ko âm,ta có:

x^2+y^2>=2xy

y^2+16>=8y

x^2+16>=8y

suy ra 2(x^2+y^2+16)>=2xy+8x+8y

suy ra x^2+y^2+16>=xy+4x+4y

a) Xét \(x^2-4x+4=\left(x-2\right)^2\ge0\)

<=> \(x^2-4x\ge-4>-5\)

b) \(2x^2+4y^2-4x-4xy+5\)

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

= \(\left(x-2\right)^2+\left(x-2y\right)^2+1\ge1>0\)

Bài 1:

a, \(A=4x^2+4x+1\)

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

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

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

Với mọi giá trị của \(x\in R\) ta có:

\(\left(2x+1\right)^2\ge0\)

Hay \(A\ge0\) với mọi giá trị của \(x\in R\).

Để \(A=0\)thì \(\left(2x+1\right)^2=0\Rightarrow2x=-1\Rightarrow x=\dfrac{-1}{2}\)

Vậy.....

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

\(B=\left[\left(x-1\right).\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

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

\(B=\left(x^2+5x+6\right)\left(x^2+5x+6\right)\)

\(B=\left(x^2+5x+6\right)^2\)

\(B=\left(x^2+2,5x+2,5x+6,25-0,25\right)^2\)

\(B=\left[\left(x+2,5\right)^2-0,25\right]^2\)

Với mọi giá trị của \(x\in R\) ta có:

\(\left(x+2,5\right)^2\ge0\Rightarrow\left(x+2,5\right)^2-0,25\ge-0,25\)

\(\Rightarrow\left[\left(x+2,5\right)^2-0,25\right]^2\ge0,0625\)

Hay \(B\ge0,0625\) với mọi giá trị của \(x\in R\).

Để \(B=0,0625\) thì \(\left[\left(x+2,5\right)^2-0,25\right]^2=0,0625\)

\(\Rightarrow\left(x+2,5\right)^2-0,25=0,25\)

\(\Rightarrow x+2,5=0\Rightarrow x=-2,5\)

Vậy.......

Câu c làm tương tự!! Chúc bạn học tốt!!!

\(A=4x^2+4x+1=\left(2x+1\right)^2\ge0\)

Vậy GTNN của A là 0 khi \(\left(2x+1\right)^2=0\Rightarrow2x+1=0\Rightarrow x=\dfrac{-1}{2}\)

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

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

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\) \(=\left(x^2+5x\right)^2-36\ge-36\)

Vậy GTNN của B là -36 khi \(\left(x^2+5x\right)^2=0\Rightarrow x\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\) \(C=x^2-2x+y^2-4y+7=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+3=\left(x-1\right)^2+\left(y-2\right)^2+3\ge3\)

Vậy GTNN của C là 3 khi \(\left[{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(A=5-8x-x^2=21-\left(16+8x+x^2\right)=21-\left(4+x\right)^2\le21\)Vậy GTLN của A là 21 khi \(4-x=0\Rightarrow x=4\)