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

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

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

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

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

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

\(C=-\left(x^2+4x+4\right)-\left(y^2-8y+16\right)+22\\ =-\left(x^2+2x.2+2^2\right)-\left(y^2-2.y.4+4^2\right)+22\\ =-\left(x+2\right)^2-\left(y-4\right)^2+22\\ Vậy:max_C=22.khi.x=-2.và.y=4\)

a: Ta có: \(A=2x^2-8x+1\)

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

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

\(=2\left(x-2\right)^2-7\ge-7\forall x\)

Dấu '=' xảy ra khi x=2

bạn làm rõ ra dc ko mik ko hiểu

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 

Bạn xem lại đề, biểu thức này ko có min max gì hết