Tim x,y sao cho: \(\sqrt{x+y+2}=\sqrt{x}+\sqrt{y}-\sqrt{2}\)
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Tim x, y sao cho: \(\sqrt{x+y-2}=\sqrt{x}+\sqrt{y}-\sqrt{2}\)
Cho x,y,z>0 va xyz=1. Tim Min cua \(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\)
cho các số dương x,y thỏa mãn\(\left(x\sqrt{x}+y\sqrt{y}\right)-3\left(x+y\right)+4\left(\sqrt{x}+\sqrt{y}\right)-4=0\)
tim ma cua M=\(\frac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
pt đã cho <=>\(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)-2\left(x+y\right)-\left(x+y+2\sqrt{xy}\right)+2\sqrt{xy}+4\left(\sqrt{x}+\sqrt{y}\right)-4=0\)
<=>\(\left(\sqrt{x}+\sqrt{y}\right)\left(x+y\right)-\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)-2\left(x+y\right)+2\sqrt{xy}-\left(\sqrt{x}+\sqrt{y}-2\right)^2=0\)
<=>\(\left(\sqrt{x}+\sqrt{y}-2\right)\left(x+y-\sqrt{xy}-\sqrt{x}-\sqrt{y}+2\right)=0\)
<=>\(\orbr{\begin{cases}\sqrt{x}+\sqrt{y}=2\\x+y-\sqrt{xy}-\sqrt{x}-\sqrt{y}+2=0\end{cases}}\)
th2: nhân cả hai vế với 2 ta được
\(\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+2>0\)
=>th2 vô nghiệm
do đó M=\(\sqrt{xy}\)
áp dụng bdt cô si ta có \(\sqrt{x}+\sqrt{y}>=2\sqrt{\sqrt{xy}}\)
<=>1>=\(\sqrt{\sqrt{xy}}\)(do \(\sqrt{x}+\sqrt{y}=2\))
<=>\(\sqrt{xy}< =1\)
<=>M<=1
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tim x, y thuoc z sao cho \(\sqrt{x}+\sqrt{y}=\sqrt{2012}\)
tìm x và y sao cho :\(\sqrt{x}+y-2=\sqrt{x}+\sqrt{y}-\sqrt{2}\)
tim x,y,z biet \(\sqrt{\left(x-\sqrt{5}\right)^2}+\sqrt{\left(y+\sqrt{3}\right)^2}+\left|x-y-z\right|\)
Tim x,y,z :1)left(2sqrt{x}-3right).left(2+sqrt{x}right)+602)sqrt{x-1+2sqrt{x-2}}+sqrt{x-1-2sqrt{x-2}}03)sqrt{x^2-4}-2sqrt{x-2}04)sqrt{x}+sqrt{y-1}+sqrt{z-2}frac{1}{2}.left(x+y+zright)5) xy xsqrt{y-1}+ysqrt{x-1}6)xsqrt{y-1}+2ysqrt{x-1}frac{3xy}{2}
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Tim x,y,z :
1)\(\left(2\sqrt{x}-3\right).\left(2+\sqrt{x}\right)+6=0\)
2)\(\sqrt{x-1+2\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}=0\)
3)\(\sqrt{x^2-4}-2\sqrt{x-2}=0\)
4)\(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}.\left(x+y+z\right)\)
5) xy =\(x\sqrt{y-1}+y\sqrt{x-1}\)
6)\(x\sqrt{y-1}+2y\sqrt{x-1}=\frac{3xy}{2}\)
a)\(\left(2\sqrt{x}-3\right)\left(2+\sqrt{x}\right)+6=0\)
\(\Leftrightarrow4\sqrt{x}+2x-6-3\sqrt{x}+6=0\)
\(\Leftrightarrow2x-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=0\\2\sqrt{x}-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{4}\end{array}\right.\)
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Tìm x và y sao cho:
\(\sqrt{x+y-2}=\sqrt{x}+\sqrt{y}-\sqrt{2}\)
Lời giải:
ĐK: \(x,y\geq 0; x+y\geq 2\)
Bình phương 2 vế thu được:
\(x+y-2=x+y+2+2\sqrt{xy}-2\sqrt{2x}-2\sqrt{2y}\)
\(\Leftrightarrow -2=2+2\sqrt{xy}-2\sqrt{2x}-2\sqrt{2y}\)
\(\Leftrightarrow 4+2\sqrt{xy}=2\sqrt{2x}+2\sqrt{2y}\)
\(\Leftrightarrow \sqrt{2}(\sqrt{x}+\sqrt{y})-2-\sqrt{xy}=0\)
\(\Leftrightarrow \sqrt{x}(\sqrt{2}-\sqrt{y})+\sqrt{2}(\sqrt{y}-\sqrt{2})=0\)
\(\Leftrightarrow (\sqrt{2}-\sqrt{y})(\sqrt{x}-\sqrt{2})=0\)
\(\Rightarrow \left[\begin{matrix} \sqrt{2}-\sqrt{y}=0\rightarrow y=2\\ \sqrt{x}-\sqrt{2}=0\rightarrow x=2\end{matrix}\right.\)
Vậy \((x,y)=(2,y)\) với $y\geq 0$ bất kỳ hoặc \((x,y)=(x,2)\) với $x\geq 0$ bất kỳ.
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Cho hàm số \(y=\sqrt{x+\sqrt{x^2+1}}\). Tính đạo gàm của hàm số.
A. \(y'=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}}\)
B. \(y'=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^2+1}}\)
C. \(y'=\dfrac{\sqrt{x^2+1}}{2\sqrt{\sqrt{x+\sqrt{x^2+1}}}}\)
D. \(y'=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)
\(y'=\dfrac{\left(x+\sqrt{x^2+1}\right)'}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{1+\dfrac{x}{\sqrt{x^2+1}}}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}.\sqrt{x+\sqrt{x^2+1}}}\)
\(=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)
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