giai pt \(x^4+\sqrt{x^2+2013}=2013\)
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Giải pt
\(\sqrt{2x+\frac{2013-1}{\sqrt{2-x^2}}}-\sqrt[3]{2014-\frac{2013-1}{\sqrt{2-x^2}}}=\sqrt{x+2013}-\sqrt[3]{x+1}\)
Giai pt sau:x-1/2013+x-2/2012+x-3/2011=x-4/2010+x-5/2009+x-6/2008
=> 3x-(1/2013+2/2012+3/2011)=3x-(4/2010+5/2009+6/2008)=>6x=-4/2010-5/2009-6/2008+1/2013+2/2012+3/2011 =>x=... làm tiếp đi bạn
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Giải pt nghiệm nguyên:
\(\dfrac{\sqrt{x-2011}-1}{x-2011} + \dfrac{\sqrt{y-2012}-1}{y-2012}+ \dfrac{\sqrt{z-2013}-1}{z-2013}= \dfrac{3}{4}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2011\\y\ge2012\\z\ge2013\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-2011}\ge0\\b=\sqrt{y-2012}\ge0\\c=\sqrt{z-2013}\ge0\end{matrix}\right.\) ta có :
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}+\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow a=b=c=2\Leftrightarrow\left\{{}\begin{matrix}x=2015\\y=2016\\z=2017\end{matrix}\right.\)
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giai pt sau:|x-2013|^5+|x-2015|^7=1
Giair pt \(\sqrt{2015-x}+\sqrt{x-2013}=x^2-4028x+4056198\)
Áp dụng bất đẳng thức bunyakovsky:
\(VT^2=\left(\sqrt{2015-x}+\sqrt{x-2013}\right)^2\le2\left(2015-x+x-2013\right)=4\)
\(\Rightarrow VT\le2\)
lại có \(VF=x^2-4028x+4056198=\left(x-2014\right)^2+2\ge2\)
do đó VT=VF khi x=2014
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Nghiệm nhỏ nhất của pt\(\frac{1}{2\sqrt{x}-2014}+\frac{1}{3\sqrt{x}+2013}=\frac{1}{2015-4\sqrt{x}}+\frac{1}{9\sqrt{x}-2016}\)
giai pt : a) \(\sqrt{\frac{2x+2}{x+2}}-\frac{\sqrt{x+2}}{\sqrt{2x+2}}=\frac{7}{12}\)
b) \(\frac{\sqrt{x-2012}-1}{x-2012}+\frac{\sqrt{y-2013}-1}{y-2013}+\frac{\sqrt{z-2014}-1}{z-2014}=\frac{3}{4}\)
c) \(3x^2+21x+18+2\sqrt{x^2+7x+7}\)= 2
Cho \(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\) Chứng minh : \(x^{2013}+y^{2013}=0\)
Cho \(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\) Chứng minh : \(x^{2013}+y^{2013=0}\)
Ta có:
\(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\\ \Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\\ \Leftrightarrow y+\sqrt{y^2+2013}=\sqrt{x^2+2013}-x\left(1\right)\)
Tương tự: \(x+\sqrt{x^2+2013}=\sqrt{y^2+2013}-y\left(2\right)\)
Do đó: 2x=-2y
Suy ra: x=-y
Do đó:
\(x^{2013}+y^{2013}=\left(-y\right)^{2013}+y^{2013}=0\left(ĐPCM\right)\)
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