\(A=2\left(x^2+y^2\right)+\left(8y^2+\dfrac{1}{2}z^2\right)+\left(8x^2+\dfrac{1}{2}z^2\right)\ge2.2\sqrt{x^2y^2}+2\sqrt{8x^2.\dfrac{1}{2}z^2}+2.\sqrt{8x^2.\dfrac{1}{2}z^2}=4\left(xy+yz+zx\right)=4\)

\(A_{min}=4\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{3};\dfrac{1}{3};\dfrac{4}{3}\right)\)

Có: \(A=16xy+\dfrac{1}{xy}-15xy\)

Áp dụng bdt Co-si, ta có:

\(16xy+\dfrac{1}{xy}\ge2\sqrt{16xy.\dfrac{1}{xy}}=8\)

Có \(x+y\ge2\sqrt{xy}< =>xy\le\dfrac{1}{4}\)

=> A \(\ge8-15.\dfrac{1}{4}=\dfrac{17}{4}\)

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

Ta có

\(A=\frac{1}{x}+\frac{1}{\sqrt{xy}}=\frac{1}{\sqrt{xx}}+\frac{1}{\sqrt{xy}}\)

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

Vậy GTNH là 2 đạt được khi x = y = 1

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

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

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

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 \(\Leftrightarrow x=y=\frac{1}{2}\)

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

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

Đặt \(\dfrac{y}{x}=a\ge4\)

\(Q=\dfrac{2a^2-2a+1}{a^2+a}=\dfrac{2a^2-2a+1}{a^2+a}-\dfrac{5}{4}+\dfrac{5}{4}=\dfrac{\left(a-4\right)\left(3a-1\right)}{4\left(a^2+1\right)}+\dfrac{5}{4}\ge\dfrac{5}{4}\)

\(Q_{min}=\dfrac{5}{4}\) khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

Đặt `(x+y)/sqrt{xy}=a(a>0)`

`P=a+1/a`

`=a+4/a-3/a`

Áp dụng BĐT cosi:

`a+4/a>=4`

`x+y>=2sqrt{xy}<=>sqrt{xy}/(x+y)<=1/2`

`<=>1/a<=1/2`

`<=>3/a<=3/2`

`<=>P>=4-3/2=8/2`

Dấu "=" `<=>x=y=1.`

Áp dụng AM-GM

\(\dfrac{x+y}{\sqrt{xy}}\ge\dfrac{2\sqrt{xy}}{\sqrt{xy}}=2\)

\(P=\dfrac{\sqrt{xy}}{x+y}+\dfrac{x+y}{4\sqrt{xy}}+\dfrac{3}{4}.\dfrac{x+y}{\sqrt{xy}}\)

Áp dụng AM-GM

\(P\ge2\sqrt{\dfrac{\sqrt{xy}}{x+y}.\dfrac{x+y}{4\sqrt{xy}}}+\dfrac{3}{4}.2=1+\dfrac{3}{2}=\dfrac{5}{2}\)

Dấu "=" \(\Leftrightarrow x=y\)