Bài 1:

a: \(A=x^2+2x+y^2+1\)

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

\(=\left(x+1\right)^2+y^2\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x+1=0\\ y=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=0\end{cases}\)

Bài 2:

a: \(x^2-5x+1\)

\(=x^2-5x+\frac{25}{4}-\frac{21}{4}\)

\(=\left(x-\frac52\right)^2-\frac{21}{4}\ge-\frac{21}{4}\forall x\)

=>\(\frac{3}{x^2-5x+1}\le3:\frac{-21}{4}=-\frac47\forall x\)

=>\(A=-\frac{3}{x^2-5x+1}\ge\frac47\forall x\)

Dấu '=' xảy ra khi \(x-\frac52=0\)

=>\(x=\frac52\)

b: \(A=\frac{6}{-x^2+2x-3}=\frac{-6}{x^2-2x+3}\)

\(=-\frac{6}{x^2-2x+1+2}=-\frac{6}{\left(x-1\right)^2+2}\)

Ta có: \(\left(x-1\right)^2+2\ge2\forall x\)

=>\(\frac{6}{\left(x-1\right)^2+2}\le\frac62=3\forall x\)

=>\(-\frac{6}{\left(x-1\right)^2+2}\ge-3\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

c: \(x^2+8\ge8\forall x\)

=>\(A=\frac{2}{x^2+8}\le\frac28=\frac14\forall x\)

Dấu '=' xảy ra khi x=0

d: \(x^2+x+4\)

\(=x^2+x+\frac14+\frac{15}{4}\)

\(=\left(x+\frac12\right)^2+\frac{15}{4}\ge\frac{15}{4}\forall x\)

=>\(A=\frac{2}{x^2+x+4}\le2:\frac{15}{4}=\frac{8}{15}\forall x\)

Dấu '=' xảy ra khi \(x+\frac12=0\)

=>\(x=-\frac12\)

a: \(\left(2x-y+7\right)^{2022}>=0\forall x,y\)

\(\left|x-1\right|^{2023}>=0\forall x\)

=>\(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}>=0\forall x,y\)

mà \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}< =0\forall x,y\)

nên \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}=0\)

=>\(\left\{{}\begin{matrix}2x-y+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x+7=9\end{matrix}\right.\)

\(P=x^{2023}+\left(y-10\right)^{2023}\)

\(=1^{2023}+\left(9-10\right)^{2023}\)

=1-1

=0

c: \(\left|x-3\right|>=0\forall x\)

=>\(\left|x-3\right|+2>=2\forall x\)

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

mà \(\left|y+3\right|>=0\forall y\)

nên \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|>=4\forall x,y\)

=>\(P=\left(\left|x-3\right|+2\right)^2+\left|y-3\right|+2019>=4+2019=2023\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y-3=0

=>x=3 và y=3

a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

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

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

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

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

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

có \(\dfrac{1}{xy}=\dfrac{2}{2xy}\left(2\right)\)

từ(1)(2)=>A=\(\dfrac{1}{1-2xy}+\dfrac{2}{2xy}\ge\dfrac{\left(1+\sqrt{2}\right)^2}{1}=\left(1+\sqrt{2}\right)^2\)

=>Min A=(1+\(\sqrt{2}\))^2

cảm ơn rất nhiều