a: Ta có: \(A=x^2+3x+4\)

\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

A= -x²+2x+3

=>A= -(x²-2x+3)

=>A= -(x²-2.x.1+1+3-1)

=>A=-[(x-1)²+2]

=>A= -(x+1)²-2

Vì -(x+1)² ≤0=> A≤-2

Dấu "=" xảy ra khi

-(x+1)²=0 => x=-1

Vây A lớn nhất= -2 khi x= -1

B=x²-2x+4y²-4y+8

=> B= (x²-2x+1)+(4y²-4y+1)+6

=> B=(x-1)²+(2y+1)²+6

=> B lớn nhất=6 khi x=1 và y=-1/2

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

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

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

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

\(a,M=x^2-4x+5=\left(x-2\right)^2+5\\ \Rightarrow M\ge5\)

Dấu "=" xảy ra \(\Leftrightarrow x=2\)

\(b,N=y^2-y-3=\left(y-\dfrac{1}{2}\right)^2-\dfrac{13}{4}\\ \Rightarrow N\ge-\dfrac{13}{4} \)

Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{1}{2}\)

\(P=x^2+y^2-4x+y+7=\left(x-2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\\ \Rightarrow P\ge\dfrac{11}{4}\)

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

a: M=x^2-4x+4+1

=(x-2)^2+1>=1

Dấu = xảy ra khi x=2

b: N=y^2-y+1/4-13/4

=(y-1/2)^2-13/4>=-13/4

Dấu = xảy ra khi y=1/2

c: P=x^2-4x+4+y^2+y+1/4+11/4

=(x-2)^2+(y+1/2)^2+11/4>=11/4

Dấu = xảy ra khi x=2 và y=-1/2

\(B=y^2-y+1\)

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

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

Ta thấy: \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\)

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

Dấu \("="\) xảy ra \(\Leftrightarrow y-\dfrac{1}{2}=0\Leftrightarrow y=\dfrac{1}{2}\)

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

\(---\)

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

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

\(=\left(x^2-2\cdot x\cdot2+2^2\right)+\left[y^2-2\cdot y\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{3}{4}\)

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

Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)

              \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2\ge0\forall x;y\)

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

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

Vậy \(C_{min}=\dfrac{3}{4}\) khi \(x=2;y=\dfrac{1}{2}\).

\(Toru\)

\(B=y^2-y+1\)

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

Vì \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\Rightarrow B\ge\dfrac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{1}{2}\)

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

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

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

Vì \(\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2\ge0\forall x,y\)

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

a) \(2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{4}\)

b) \(5x-x^2+4=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{5}{2}\)

c) \(x^2+5y^2-2xy+4y+3=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

\(ĐTXR\Leftrightarrow\)\(x=y=-\dfrac{1}{2}\)

b: ta có: \(-x^2+5x+4\)

\(=-\left(x^2-5x-4\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}\right)\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

Áp dụng Bunyakovsky, ta có :

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

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

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

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