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Áp dụng BĐT cosi:
`1/x^2+1/y^2>=2/(xy)`
`<=>2>=2/(xy)`
`<=>1>=1/(xy)`
`<=>xy>=1`
Dấu "=" xảy ra khi `x=y=1`
SỬa đề: x^3-xy^2
\(A=\left(\dfrac{x\left(x-y\right)}{y\left(x+y\right)}+\dfrac{x^2-y}{x\left(x+y\right)}\right):\left(\dfrac{y^2}{x\left(x^2-y^2\right)}+\dfrac{1}{x-y}\right)\)
\(=\left(\dfrac{x^2\left(x-y\right)+y\left(x^2-y\right)}{xy\left(x+y\right)}\right):\left(\dfrac{y^2}{x\left(x-y\right)\left(x+y\right)}+\dfrac{x\left(x+y\right)}{x\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^3-x^2y+x^2y-y^3}{xy\left(x+y\right)}:\dfrac{y^2+x^2+xy}{x\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\cdot\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2+xy+y^2}=\dfrac{\left(x-y\right)^2}{y}\)
Để A>0 thì y>0
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy})(x^2+y^2+2xy)\geq (1+1+2)^2=16$
$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\geq \frac{16}{(x+y)^2}=16$
Áp dụng BĐT AM-GM:
$xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}$
$\Rightarrow \frac{2}{xy}\geq 8$
Cộng 2 BĐT trên lại:
$P\geq 16+8=24$
Vậy $P_{\min}=24$ khi $x=y=\frac{1}{2}$
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy})(x^2+y^2+2xy)\geq (1+1+2)^2=16$
$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\geq \frac{16}{(x+y)^2}=16$
Áp dụng BĐT AM-GM:
$xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}$
$\Rightarrow \frac{2}{xy}\geq 8$
Cộng 2 BĐT trên lại:
$P\geq 16+8=24$
Vậy $P_{\min}=24$ khi $x=y=\frac{1}{2}$
A=(xy2+xy−x−yx2+xy) :
A=( \(\dfrac{x}{y\left(x+y\right)}\) - \(\dfrac{x-y}{x\left(x+y\right)}\)) : (\(\dfrac{y^2}{x\left(x-y\right)\left(x+y\right)}\)+\(\dfrac{1}{x+y}\)) : \(\dfrac{x}{y}\)
A=\(\dfrac{x^2-y\left(x-y\right)}{xy\left(x+y\right)}\) : \(\dfrac{y^2+x\left(x-y\right)}{x\left(x-y\right)\left(x+y\right)}\) : \(\dfrac{x}{y}\)
A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\) : \(\dfrac{y^2-xy+x^2}{x\left(x-y\right)\left(x+y\right)}\):\(\dfrac{x}{y}\)
A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\). \(\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2-xy+y^2}\):\(\dfrac{x}{y}\)
A = \(\dfrac{x-y}{y}\) : \(\dfrac{x}{y}\)
A = \(\dfrac{x-y}{x}\)
A= 1 - \(\dfrac{y}{x}\)>1
=> y/x <0
=> xy<0 , x+y khác 0
\(6xy=x+y\ge2\sqrt[]{xy}\Rightarrow\sqrt{xy}\ge\dfrac{1}{3}\Rightarrow xy\ge\dfrac{1}{9}\Rightarrow\dfrac{1}{xy}\le9\)
\(M=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{xy+x}{1-xy}+1}{1+\dfrac{xy+x}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{x+1}{1-xy}}{\dfrac{x+1}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{1}{1-xy}+\dfrac{1}{1+xy}}{\dfrac{1}{1-xy}-\dfrac{1}{1+xy}}\)
\(M=\dfrac{1+xy+1-xy}{1+xy-1+xy}=\dfrac{2}{2xy}=\dfrac{1}{xy}\le9\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{3}\)