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\(6xy=x+y\ge2\sqrt[]{xy}\Rightarrow\sqrt{xy}\ge\dfrac{1}{3}\Rightarrow xy\ge\dfrac{1}{9}\Rightarrow\dfrac{1}{xy}\le9\)
\(M=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{xy+x}{1-xy}+1}{1+\dfrac{xy+x}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{x+1}{1-xy}}{\dfrac{x+1}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{1}{1-xy}+\dfrac{1}{1+xy}}{\dfrac{1}{1-xy}-\dfrac{1}{1+xy}}\)
\(M=\dfrac{1+xy+1-xy}{1+xy-1+xy}=\dfrac{2}{2xy}=\dfrac{1}{xy}\le9\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{3}\)
Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:
\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).
Không mất tính tổng quát, giả sử x + y = 0
\(\Leftrightarrow x=-y\)
\(\Leftrightarrow x^3=-y^3\).
Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).
Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).
Áp dụng cosi
`1/x^2+1/y^2>=2/(xy)`
`=>1/2>=2/(xy)`
`=>xy>=4`
Aps dụng cosi
`=>x+y>=2\sqrt{xy}=2.2=4`
Dấu "=" xảy ra khi `x=y=4`
Có : \(\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge2\sqrt{\dfrac{1}{x^2}\cdot\dfrac{1}{y^2}}=\dfrac{2}{xy}\)
\(\Rightarrow xy\ge4\)
Ta có : \(A=x+y\ge2\sqrt{xy}=2\sqrt{4}=4\)
Dấu "=" xảy ra khi \(x=y=2\)
Vậy min A = 4 khi $x=y=2$
\(P=\dfrac{x+2y}{2xy}+\dfrac{1}{x+2y}=\dfrac{x+2y}{4}+\dfrac{1}{x+2y}\)
\(P=\dfrac{x+2y}{16}+\dfrac{1}{x+2y}+\dfrac{3\left(x+2y\right)}{16}\)
\(P\ge2\sqrt{\dfrac{x+2y}{16\left(x+2y\right)}}+\dfrac{3}{16}.2\sqrt{2xy}=\dfrac{5}{4}\)
\(P_{min}=\dfrac{5}{4}\) khi \(\left(x;y\right)=\left(2;1\right)\)
Ta có:
\(M=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}=\frac{1}{16x}+\frac{4}{16y}+\frac{16}{16z}\)
\(\ge\frac{\left(1+2+4\right)^2}{16\left(x+y+z\right)}=\frac{49}{16}\)
Dấu bằng xảy ra khi
\(\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}\)
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`
\(\dfrac{1}{x}+\dfrac{1}{y}=-1\Rightarrow\dfrac{x+y}{xy}=-1\Rightarrow x+y=-xy\)
\(\Rightarrow\left(x+y\right)^3=-x^3y^3\)
\(S=\dfrac{y}{x^2}+\dfrac{x}{y^2}+xy=\dfrac{x^3+y^3+x^3y^3}{x^2y^2}=\dfrac{x^3+y^3-\left(x+y\right)^3}{x^2y^2}=\dfrac{-3xy\left(x+y\right)}{x^2y^2}=\dfrac{-3xy.\left(-xy\right)}{x^2y^2}=\dfrac{3x^2y^2}{x^2y^2}=3\)
We have \(\dfrac{1}{x}+\dfrac{1}{y}=-1\Leftrightarrow\dfrac{x+y}{xy}=-1\Leftrightarrow x+y=-xy\)
Base on this, we have \(S=\dfrac{y}{x^2}+\dfrac{x}{y^2}+xy\)\(=\dfrac{x^3+y^3}{x^2y^2}+xy\) \(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x^2y^2}+xy\) \(=\dfrac{-xy\left[\left(x+y\right)^2-3xy\right]}{x^2y^2}+xy\) \(=\dfrac{-xy\left[\left(-xy\right)^2-3xy\right]}{x^2y^2}+xy\)\(=\dfrac{-xy\left(x^2y^2-3xy\right)}{x^2y^2}+xy\) \(=\dfrac{-x^2y^2\left(xy-3\right)}{x^2y^2}+xy\) \(=-\left(xy-3\right)+xy\) \(=3\)
In conlusion, with \(x,y\inℝ\) and \(x,y\ne0\) such that \(\dfrac{1}{x}+\dfrac{1}{y}=-1\), we have \(S=3\)