\(a,=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(b,=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

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

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)

Ta có

ĐÁP ÁN C

Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

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

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

\(=3\left(x-2\right)^2-12>=-12\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

\(A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+4\\ A=\left(x-y\right)^2+\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=1\end{matrix}\right.\Leftrightarrow x=y=1\)

\(A=x^2+y^2+\left(\dfrac{1}{2}\right)^2-2xy+2.\dfrac{1}{2}x-2.\dfrac{1}{2}.y+\dfrac{3}{4}\)

\(A=\left(x-y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(A_{min}=\dfrac{3}{4}\) khi \(x-y+\dfrac{1}{2}=0\)

\(A=3x^2+y^2+2xy+4x\)

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

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

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

Dễ thấy: \(2\left(x+1\right)^2+\left(x+y\right)^2\ge0\)

\(\Rightarrow2\left(x+1\right)^2+\left(x+y\right)^2-2\ge-2\)

Xảy ra khi \(\hept{\begin{cases}x+1=0\\x+y=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-1\\x=-y\end{cases}}\Rightarrow x=-y=-1\)

A=-x^2+2xy-y^2-x^2+4x-4-36

=-(x-y)^2-(x-2)^2-36<=-36

Dấu = xảy ra khi x=y=2

a.

\(A=x^2-4x+4+2=\left(x-2\right)^2+2\ge2\)

GTNN của A đạt 2 khi và chỉ khi \(x=2\)

b.

\(B=y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+\dfrac{3}{4}=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

GTNN của B đạt \(\dfrac{3}{4}\) khi và chỉ khi \(y=\dfrac{1}{2}\)

c.

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

GTNN của C đạt \(\dfrac{3}{4}\) khi và chỉ khi \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

a) \(A=x^2-4x+6\)

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

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

Mà: \(\left(x-2\right)^2\ge0\forall x\) nên \(A=\left(x-2\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra:

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

\(\Leftrightarrow x=2\)

Vậy: \(A_{min}=2\) khi \(x=2\)

b) \(B=y^2-y+1\)

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

\(B=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Mà: \(\left(y-\dfrac{1}{2}\right)^2\ge\forall x\) nên \(B=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu "=" xảy ra:

\(\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\Leftrightarrow y-\dfrac{1}{2}=0\)

\(\Leftrightarrow y=\dfrac{1}{2}\)

Vậy \(B_{min}=\dfrac{3}{4}\) khi \(y=\dfrac{1}{2}\)

c) \(C=x^2-4x+y^2-y+5\)

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

\(C=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

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

\(C=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu "=" xảy ra:

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

Vậy: \(C_{min}=\dfrac{3}{4}\) khi \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Rightarrow4A=4x^2+4y^2+4z^2-4yz-16x-12y+8108=4x^2-16x+16+3y^2+12y+12+y^2-4yz+4z^2+8080=4\left(x-2\right)^2+3\left(y+2\right)^2+\left(y-2z\right)^2+8080\)

Vì \(4\left(x-2\right)^2\ge0\)

    \(3\left(y+2\right)^2\ge0\)

     \(\left(y-2z\right)^2\ge0\)

\(\Rightarrow4A\ge8080\Rightarrow A\ge2020\)

\(ĐTXR\Leftrightarrow x=2,y=-2,z=-1\)

sao lại \(2x^x\)