\(P=x^2+2xy+4x+4y+y^2+5\)

  \(=\left(x^2+2xy+y^2\right)+4\left(x+y\right)+5\)

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

  \(=\left(x+y+2\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow x+y+2=0\)

Vậy với x + y + 2 = 0 thì P_min = 1

p = x.x + 2.x.y+ 4.x+4.y+ y.2+5

=> P= x.(x+2+y+4)+y.(4+2) +5

mà giá trị nhỏ nhất là gì ạ?

\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}=\frac{5x^2-y^2}{xy}-\frac{3x^2-2xy}{xy}\)

\(=\frac{5x^2-y^2-3x^2+2xy}{xy}=\frac{2x^2+2xy-y^2}{xy}\)

\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\left(Đk:x;y\ne0\right)\)

\(=\frac{5x^2-y^2}{xy}-\frac{3x^2-2xy}{xy}=\frac{5x^2-y^2-3x^2+2xy}{xy}\)

\(=\frac{2x^2+2xy-y^2}{xy}\)\(=\frac{x^2+\left(x+y\right)^2}{xy}\)

Đặt A = 4x² - 6x + 23

A = (2x)² - 2.2.1,5.x + 2,25 + 20,75

\(A=\left(2x-1,5\right)^2+20,75\ge20,75.\)

Để A nhỏ nhất

=> (2x-1,5)² + 20,75 = 20,75

=> (2x-1,5)² = 0

2x = 1,5

x = 0,75

KL: \(Min_A=20,75\)tại x = 0,75

\(4x^2-6x+23\)

\(=\left(2x\right)^2-2.\frac{3}{2}.2x+\frac{9}{4}+\frac{83}{4}\)

\(=\left(2x-\frac{3}{2}\right)^2+\frac{83}{4}\ge\frac{83}{4}\)

Dấu "=" <=> 2x = 3/2 

             <=> x = 3/4

Vậy

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\) (*)

Ta có: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\)

\(=\frac{x}{z}+\frac{y}{z}+\frac{x}{y}+\frac{x}{y}+\frac{y}{x}+\frac{z}{x}\)

\(=\left(\frac{x}{z}+\frac{x}{y}\right)+\left(\frac{y}{x}+\frac{y}{z}\right)+\left(\frac{z}{x}+\frac{z}{y}\right)\)

\(=x\left(\frac{1}{z}+\frac{1}{y}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)

Thay (*) vào,ta có : \(A=x.\left(\frac{-1}{x}\right)+y.\left(-\frac{1}{y}\right)+z.\left(-\frac{1}{z}\right)=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)

a) \(a^2+b^2+1\ge ab+a+b\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra <=> a=b=1.

b) \(a^2-2a+6b+b^2=-10\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+6b+9\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\). Mà \(\left(a-1\right)^2\ge0;\left(b+3\right)^2\ge0\forall a;b\)

Nên \(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=-3\end{cases}}}\). KL: ...