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ta co:\(y^2\sqrt{x-2}-2y+\sqrt[]{x-2}=0\)
xét denta:\(\Delta=b^2-4ac=4-4.\left(x-2\right)=4\left(3-x\right)\)
để có y thỏa mãn => denta >=0
=>\(3>=x\)
=>dpcm
Áp dụng bdt cosi-schwar cho 3 số (\(\left(am+bn+cp\right)^2\le\left(a^2+b^2+c^2\right)\)\(\left(m^2+n^2+p^2\right)\)
với a=x,b=y\(\sqrt{2}\);c=z\(\sqrt{5}\); m=\(\sqrt{11-2y^2},n=\sqrt{3-5z^2}\),\(p=\sqrt{2-x^2}\)
82\(\le\left(x^2+2y^2+5z^2\right)\left(11-2y^2+3-5z^2+1-x^2\right)\) <=>64\(\le P\left(16-P\right)\)
<=>P2-16P+64\(\le0< =>\left(P-8\right)^2\le0\) <=>P=8
\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-1=0\\\sqrt{y-1}-1=0\Leftrightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\\\sqrt{z-2}-1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-1=0\\\sqrt{y-1}-1=0\Leftrightarrow\\\sqrt{z-2}-1=0\end{cases}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}}\)
vậy \(S=x+y=1+2=3\)
\(x^3+y^3=2x^2y^2\Rightarrow\)\(\left(x^3+y^3\right)^2=4x^4y^4\Rightarrow x^6+2x^3y^3+y^6=4x^4y^4\)\(\Rightarrow x^6+2x^3y^3+y^6-4x^3y^3=4x^4y^4-4x^3y^3\)\(\Rightarrow\left(x^3-y^3\right)^2=4x^4y^4\left(1-\frac{1}{xy}\right)\)
\(\Rightarrow\sqrt{1-\frac{1}{xy}}=\)\(\frac{\left|x^3-y^3\right|}{2x^2y^2}\)mà x:y hữu tie suy ra điều phải cm
Cái bài này bạn làm ra chưa:
\(\sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}\ge2\left(\sqrt{\frac{c}{a+b}}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}\right)\)
1. Ta có: \(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Rightarrow\left(x+y+z\right)^2=\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=xy+yz+zx+2y\sqrt{xz}+2z\sqrt{xy}+2x\sqrt{yz}\)
\(\Leftrightarrow x^2+y^2+z^2+xy+yz+zx-2y\sqrt{xz}-2z\sqrt{xy}-2x\sqrt{yz}=0\)
\(\Leftrightarrow\left(x-\sqrt{yz}\right)^2+\left(y-\sqrt{xz}\right)^2+\left(z-\sqrt{xy}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{yz}\\y=\sqrt{xz}\\z=\sqrt{xy}\end{matrix}\right.\)
\(\Rightarrow x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\Rightarrow x=y=z\)
Bài 1:
\(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)
\(\Leftrightarrow x+y+z-\sqrt{xy}-\sqrt{yz}-\sqrt{xz}=0\)
\(\Leftrightarrow 2x+2y+2z-2\sqrt{xy}-2\sqrt{yz}-2\sqrt{xz}=0\)
\(\Leftrightarrow (x+y-2\sqrt{xy})+(y+z-2\sqrt{yz})+(z+x-2\sqrt{xz})=0\)
\(\Leftrightarrow (\sqrt{x}-\sqrt{y})^2+(\sqrt{y}-\sqrt{z})^2+(\sqrt{z}-\sqrt{x})^2=0\)
Vì \( (\sqrt{x}-\sqrt{y})^2;(\sqrt{y}-\sqrt{z})^2;(\sqrt{z}-\sqrt{x})^2\geq 0, \forall x,y,z>0\) nên để tổng của chúng bằng $0$ thì:
\( (\sqrt{x}-\sqrt{y})^2=(\sqrt{y}-\sqrt{z})^2=(\sqrt{z}-\sqrt{x})^2=0\)
\(\Rightarrow x=y=z\) (đpcm)
Đặt \(\sqrt{x-2}=a\ge0\)
\(\Rightarrow ay^2-2y+a=0\)
\(\Delta'=1-a^2\ge0\Rightarrow\left|a\right|\le1\Rightarrow0\le a\le1\)
\(\Rightarrow\sqrt{x-2}\le1\Rightarrow x\le3\Rightarrow x^3\le27\)