ĐKXĐ: \(x\ge0;y\ge1;z\ge2\)
\(x-2\sqrt{x}+1+y-1-4\sqrt{y-1}+4+z-2-6\sqrt{z-2}+9=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-2\right)^2+\left(\sqrt{z-2}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{y-1}=2\\\sqrt{z-2}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=5\\z=11\end{matrix}\right.\)
tìm x,y,z biết \(x+y+z=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}-8\)
Tìm x, y, z biết:
\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}=\dfrac{1}{2}\left(x+y+z\right)\)
Tìm các số x, y thỏa mãn đẳng thức:
a, \(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
b, \(\sqrt{x-2}+\sqrt{y+2009}+\sqrt{z-2010}=\dfrac{1}{2}\left(x+y+z\right)\)
Tìm x,y,z biết:
x+y+z+35 = \(2.\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)
Mấy bạn giúp mình bài này nha!
1) Tính A=(\(\sqrt{6}+\sqrt{2}\))*(\(\sqrt{3}-2\))*\(\sqrt{2+\sqrt{3}}\)
2) Cho x=4+\(\sqrt{10}\)
Tính A=\(\sqrt{3x+\sqrt{6x-1}}+\sqrt{3x-\sqrt{6x-1}}\)
3) Cho \(\sqrt{x}+\sqrt{y}-\sqrt{z}=0\)
CMR: \(\dfrac{1}{x+y-z}+\dfrac{1}{y+z-x}+\dfrac{1}{z+x-y}=0\)
4) Cho (\(\sqrt{x^2+5}+x\))*(\(\sqrt{y^2+5}+y\))=4
CMR: x+y=0
Cho 3 số x y z thỏa mãn x+y+z=xyz.Cm:\(\dfrac{\sqrt{\left(1+y^2\right)\left(1+z^2\right)}-\sqrt{1+y^2}-\sqrt{1+z^2}}{yz}+\dfrac{\sqrt{\left(1+z^2\right)\left(1+x^2\right)}-\sqrt{1+z^2}-\sqrt{1+x^2}}{zx}+\dfrac{\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\sqrt{1+x^2}-\sqrt{1+z^2}}{yz}=0\)
Cho 3 số dương x,y,z thỏa mãn x + y + z = xyz. Cmr:
\(A=\frac{\sqrt{\left(1+y^2\right)\left(1+z^2\right)}-\sqrt{1+y^2}-\sqrt{1+z^2}}{yz}+\frac{\sqrt{\left(1+z^2\right)\left(1+x^2\right)}-\sqrt{1+x^2}-\sqrt{1+z^2}}{xz}+\frac{\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\sqrt{1+x^2}-\sqrt{1+y^2}}{xy}=0\)
Tìm x,y,z biết: \(\sqrt{x-1}\) + \(\sqrt{y-2}\) + \(\sqrt{z-3}\)= 6 - \(\dfrac{1}{\sqrt{x-1}}\) - \(\dfrac{1}{\sqrt{y-2}}\) - \(\dfrac{1}{\sqrt{z-3}}\) .
Giúp mình nha ^^
Cho x,y,z >0 thỏa x+y+z=\(\sqrt{2021}\)
Tìm Min:
\(P=\sqrt{\left(x+y\right)\left(y+z\right)\left(z+x\right)}.\left(\dfrac{\sqrt{y+z}}{x}+\dfrac{\sqrt{z+x}}{y}+\dfrac{\sqrt{x+y}}{z}\right)\)
