Ai giúp mik vs

Huhu ai giúp vs

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

b, ta có : \(x+y=1=>2x+2y=2\)

\(B=\dfrac{1}{x^2+y^2}+\dfrac{3}{4xy}=\dfrac{4}{4x^2+4y^2}+\dfrac{6}{8xy}\)\(\ge\dfrac{\left(2+\sqrt{6}\right)^2}{\left(2x+2y\right)^2}\)

\(=\dfrac{\left(2+\sqrt{6}\right)^2}{2^2}=\dfrac{5+2\sqrt{6}}{2}\)=>\(B\ge\dfrac{5+2\sqrt{6}}{2}\)

=>\(MinB=\dfrac{5+2\sqrt{6}}{2}\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{5}{4xy}\)

\(\ge\dfrac{4}{x^2+y^2+2xy}+2\sqrt{\dfrac{1}{4xy}.4xy}+\dfrac{5}{4.\dfrac{\left(x+y\right)^2}{4}}\)

\(\ge\dfrac{4}{1^2}+2+\dfrac{5}{1^2}\) (do \(x+y\le1\))

\(=11\)

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

Vậy GTNN của A là 11.

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 \(\Leftrightarrow x=y=\frac{1}{2}\)

LOADING...

Ta có : \(x+y\ge2\sqrt{xy}\) \(\Rightarrow xy+2\sqrt{xy}\le8\) hay \(\left(\sqrt{xy}+1\right)^2\le9\)

\(\Rightarrow\sqrt{xy}+1\le3\Rightarrow xy\le4\)

Ta có : \(\left(9-xy\right)^2=\left(x+y+1\right)^2=x^2+y^2+1+2\left(x+y+xy\right)=x^2+y^2+17\)

Vì \(xy\le4\Rightarrow9-xy\ge5\Rightarrow\left(9-xy\right)^2\ge25\Leftrightarrow x^2+y^2+17\ge25\)

\(\Rightarrow A\ge8\) . Dấu "=" xảy ra khi x = y = 2

Vậy Min A = 8 tại x = y = 2

Ta có:

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

\(=\frac{1}{3}\left(x^2+4+y^2+4\right)+\frac{2}{3}\left(x^2+y^2\right)-\frac{8}{3}\)

\(\ge\frac{4}{3}\left(x+y+xy\right)-\frac{8}{3}=8\)

\(\Rightarrow P\ge8\)

Dấu = khi \(x=y=2\)

Vậy MinP=8 khi x=y=2

\(y\ge1+xy\Rightarrow1\ge\dfrac{1}{y}+x\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le4\Rightarrow\dfrac{y}{x}\ge4\)

\(G=\dfrac{x}{y}+\dfrac{y}{x}=\left(\dfrac{x}{y}+\dfrac{y}{16x}\right)+\dfrac{15}{16}.\dfrac{y}{x}\ge2\sqrt{\dfrac{xy}{16xy}}+\dfrac{15}{16}.4=\dfrac{17}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)