\(M=2x^2+2y^2-2xy-2x+2y+2\)

\(=\left[\left(x^2-2xy+y^2\right)-\frac{4}{3}\left(x-y\right)+\frac{4}{9}\right]+\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+\left(y^2+\frac{2}{3}y+\frac{1}{9}\right)+\frac{4}{3}\)

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

\(\Rightarrow M\ge\frac{2}{3}\)

\(M=x^2+y^2-xy-x+y+1\)

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

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

\(4M=\left[\left(2x-y\right)^2-2\left(2x-y\right)+1\right]+3\left(y^2+2y+1\right)-3\)

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

Mà : \(\left(2x-y-1\right)^2\ge0\forall x;y\)

\(\left(y+1\right)^2\ge0\forall y\Rightarrow3\left(y+1\right)^2\ge0\forall y\)

\(\Rightarrow4M\ge-3\)

\(\Leftrightarrow M\ge-\frac{3}{4}\)

Dấu " = " xảy ra khi :

\(\hept{\begin{cases}2x-y-1=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-y=1\\y=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)

Vậy  \(M_{Min}=-\frac{3}{4}\Leftrightarrow\left(x;y\right)=\left(0;-1\right)\)

Dùng Cô - si nha :))

Áp dụng BĐT AM - GM cho hai số dương ta có :

\(T=xy+\frac{10}{xy}=\left(10xy+\frac{10}{xy}\right)-9xy\)

\(\ge2\sqrt{10xy\cdot\frac{10}{xy}}-9\cdot\frac{\left(x+y\right)^2}{4}\)

\(=20-9\cdot1=11\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=1\)

Vậy : \(minT=11\) tại \(x=y=1\)

\(M=x^2+y^2-xy-x-y+1\)

\(\Rightarrow2M=2x^2+2y^2-2xy-2x-2y+2\)

\(\Rightarrow2M=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)\)

\(\Rightarrow2M=\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2>0\)

\(\Rightarrow M>0\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x-y=0\\x-1=0\\y-1=0\end{cases}}\Rightarrow x=y=1\)

Thay y= 1-x ta được

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

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

Đẳng thức xảy ra  \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y=1-x\end{cases}}\)  \(\Leftrightarrow x=y=\frac{1}{2}\)

Đặt \(x=1-y\)

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

\(\Leftrightarrow C=1-2y+y^2+y^2+y-y^2=y^2-y+1\)

\(\Leftrightarrow\left(y^2-2.\frac{1}{2}y+\frac{1}{4}\right)+\frac{3}{4}\Leftrightarrow\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy min C là 3/4 khi y=1/2 và x =1- 1/2= 1/2 hay x=y= 1/2

\(A=\left(x^3+y^3+xy\left(x+y\right)\right)-xy\left(x+y\right)+xy\)

=>    \(A=\left(x+y\right)\left(x^2+y^2\right)-xy.1+xy\)

=>   \(A=x^2+y^2-xy+xy\)

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

DẤU "=" XẢY RA <=>    \(x=y\). MÀ    \(x+y=1\)

=> A min \(=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\).

\(B=x^2-2x+1+x^2-6x+9\)

=>   \(B=2x^2-8x+10\)

=>   \(B=2\left(x^2-4x+4\right)+2\)

=>   \(B=2\left(x-2\right)^2+2\)

CÓ:    \(2\left(x-2\right)^2\ge0\forall x\Rightarrow2\left(x-2\right)^2+2\ge2\)

=>   \(B\ge2\)

DẤU "=" XẢY RA <=>    \(2\left(x-2\right)^2=0\Leftrightarrow x=2\)

VẬY B MIN = 2 <=>    \(x=2\)

Bài 1:

\(x^2-5x-6=0\)

\(\Leftrightarrow x^2+x-6x-6=0\)

\(\Leftrightarrow\left(x^2+x\right)-\left(6x+6\right)=0\)

\(\Leftrightarrow x\left(x+1\right)-6\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)

Vậy x=-1; x=6

Bài 2:

a) Ta có: \(x+y=10\Leftrightarrow y=10-x\) (1)

Từ (1) thay vào \(P=xy\) ta được:

\(P=x\left(10-x\right)\)

\(\Leftrightarrow P=10x-x^2\)

\(\Leftrightarrow P=-x^2+10x-5^2+5^2\)

\(\Leftrightarrow P=-\left(x^2-10x+5^2\right)+5^2\)

\(\Leftrightarrow P=-\left(x-5\right)^2+25\)

Vậy GTLN của P=25 khi \(x-5=0\Leftrightarrow x=5\)

b) \(P=x^2-5x\)

\(\Leftrightarrow P=x^2-2x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\left(\dfrac{5}{2}\right)^2\)

\(\Leftrightarrow P=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\)

Vậy GTNN của \(P=\dfrac{-25}{4}\) khi \(x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)