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\(\sqrt{x-29}+2\sqrt{y-6}+3\sqrt{z-2011}+1016=\dfrac{1}{2}\left(x+y+z\right)\)\(\Leftrightarrow2\sqrt{x-29}+4\sqrt{y-6}+6\sqrt{z-2011}+2032=x+y+z\)\(\Leftrightarrow-2\sqrt{x-29}-4\sqrt{y-6}-6\sqrt{z-2011}-2032=-x-y-z\)\(\Leftrightarrow(x-29-2\sqrt{x-29}+1)+(y-6-2\cdot2\sqrt{y-6}+2^2)+(z-2011-2\cdot3\sqrt{z-2011}+3^2)=0\)\(\Leftrightarrow\left(\sqrt{x-29}-1\right)^2+\left(\sqrt{y-6}-2\right)^2+\left(\sqrt{z-2011}-3\right)^2=0\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-29}-1=0\\\sqrt{y-6}-2=0\\\sqrt{z-2011}-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-29}=1\\\sqrt{y-6}=2\\\sqrt{z-2011}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-29=1\\y-6=4\\z-2011=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=10\\z=2020\end{matrix}\right.\)
Vậy : ......................
Điều kiện xác định : \(x\ge0\),\(y\ge1\),\(z\ge2\)
\(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
Mà \(\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2\ge0\)
Đẳng thức xảy ra khi \(\left(\sqrt{x}-1\right)^2=\left(\sqrt{y-1}-1\right)^2=\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)
\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-1=0\\\sqrt{y-1}-1=0\Leftrightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\\\sqrt{z-2}-1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-1=0\\\sqrt{y-1}-1=0\Leftrightarrow\\\sqrt{z-2}-1=0\end{cases}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}}\)
vậy \(S=x+y=1+2=3\)
\(\frac{x+\left(\sqrt{x}-\sqrt{z}\right)^2}{y+\left(\sqrt{y}-\sqrt{z}\right)^2}=\frac{\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)^2-y+\left(\sqrt{x}-\sqrt{z}\right)^2}{\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)^2-x+\left(\sqrt{y}-\sqrt{z}\right)^2}\)
\(=\frac{\left(\sqrt{x}+2\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)+\left(\sqrt{x}-\sqrt{z}\right)^2}{\left(2\sqrt{x}+\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)+\left(\sqrt{y}-\sqrt{z}\right)^2}\)
\(=\frac{\left(\sqrt{x}-\sqrt{z}\right)\left(2\sqrt{x}+2\sqrt{y}-2\sqrt{z}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(2\sqrt{x}+2\sqrt{y}-2\sqrt{z}\right)}\)
\(=\frac{\sqrt{x}-\sqrt{z}}{\sqrt{y}-\sqrt{z}}\)
\(x+y+z=2\sqrt{x-29}+4\sqrt{y-6}+6\sqrt{z-2011}+2032\)
<=>\(\left(x-29\right)-2\sqrt{x-29\cdot}+1+\left(y-6\right)-4\sqrt{y-6}+4+\left(z-2011\right)-6\sqrt{z-2011}+9=0\)
<=>\(\left(\sqrt{x-29}-1\right)^2+\left(\sqrt{y-6}-2\right)^2+\left(\sqrt{z-2011}-3\right)^2=0\)
cho 3 cái =0 là ra
nhân 2 lên rồi rút về hằng đẳng thức là xong bạn ak cần mk giải ra ko
Ta có : \(\left(x+\sqrt{x^2+2017}\right)\left(-x+\sqrt{x^2+2017}\right)=2017\left(1\right)\)
\(\left(y+\sqrt{y^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\left(2\right)\)
nhân theo vế của ( 1 ) ; ( 2 ) , ta có :
\(2017\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017^2\)
\(\Rightarrow\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\)
rồi bạn nhân ra , kết hợp với việc nhân biểu thức ở phần trên xong cộng từng vế , cuối cùng ta đc :
\(xy+\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017\)
\(\Leftrightarrow\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017-xy\)
\(\Leftrightarrow x^2y^2+2017\left(x^2+y^2\right)+2017^2=2017^2-2\cdot2017xy+x^2y^2\)
\(\Rightarrow x^2+y^2=-2xy\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)
A = 2017
( phần trên mk lười nên không nhân ra, bạn giúp mk nhân ra nha :) )
2/ \(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{4\sqrt{x-2011}-4}{x-2011}+\frac{4\sqrt{y-2012}-4}{y-2012}+\frac{4\sqrt{z-2013}-4}{z-2013}=3\)
\(\Leftrightarrow\left(1-\frac{4\sqrt{x-2011}-4}{x-2011}\right)+\left(1-\frac{4\sqrt{y-2012}-4}{y-2012}\right)+\left(1-\frac{4\sqrt{z-2013}-4}{z-2013}\right)=0\)
\(\Leftrightarrow\left(\frac{x-2011-4\sqrt{x-2011}+4}{x-2011}\right)+\left(\frac{y-2012-4\sqrt{y-2012}+4}{y-2012}\right)+\left(\frac{z-2013-4\sqrt{z-2013}+4}{z-2013}\right)=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2011}-2\right)^2}{x-2011}+\frac{\left(\sqrt{y-2012}-2\right)^2}{y-2012}+\frac{\left(\sqrt{z-2013}-2\right)^2}{z-2013}=0\)
Dấu = xảy ra khi \(\sqrt{x-2011}=2;\sqrt{y-2012}=2;\sqrt{z-2013}=2\)
\(\Leftrightarrow x=2015;y=2016;z=2017\)
ĐKXĐ: ....
\(\Leftrightarrow2\sqrt{x-29}+4\sqrt{y-6}+6\sqrt{z-2011}+2032=x+y+z\)
\(\Leftrightarrow x-29-2\sqrt{x-29}+1+y-6-4\sqrt{y-6}+4+z-2011-6\sqrt{z-2011}+9=0\)
\(\Leftrightarrow\left(\sqrt{x-29}-1\right)^2+\left(\sqrt{y-6}-2\right)^2+\left(\sqrt{z-2011}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-29}-1=0\\\sqrt{y-6}-2=0\\\sqrt{z-2011}-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=30\\y=10\\z=2020\end{matrix}\right.\)