Giải pt:
\(x^2\left(x^2+2\right)=4-x\sqrt{2x^2+4}\)
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NL
Nguyễn Lê Phước Thịnh
CTVHS
26 tháng 3
c: ĐKXĐ: x>=1/2
Ta có: \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt2\)
=>\(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
=>\(\sqrt{2x-1+2\cdot\sqrt{2x-1}\cdot1+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
=>\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
=>\(\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|=2\)
=>\(\left|\sqrt{2x-1}-1\right|=2-\sqrt{2x-1}-1=-\sqrt{2x-1}+1=-\left(\sqrt{2x-1}-1\right)\)
=>\(\sqrt{2x-1}-1\le0\)
=>\(\sqrt{2x-1}\le1\)
=>2x-1<=1
=>2x<=2
=>x<=1
=>1/2<=x<=1
d:
ĐKXĐ: x>=-1/4
\(x+\sqrt{x+\frac12+\sqrt{x+\frac14}}=4\)
=>\(x+\sqrt{x+\frac14+2\cdot\sqrt{x+\frac14}\cdot\frac12+\frac14}=4\)
=>\(x+\sqrt{\left(\sqrt{x+\frac14}+\frac12\right)^2}=4\)
=>\(x+\sqrt{x+\frac14}+\frac12=4\)
=>\(x+\frac12+\sqrt{x+\frac14}=4\)
=>\(x+\frac14+2\cdot\sqrt{x+\frac14}\cdot\frac12+\frac14=4\)
=>\(\left(\sqrt{x+\frac14}+\frac12\right)^2=4\)
=>\(\sqrt{x+\frac14}+\frac12=2\)
=>\(\sqrt{x+\frac14}=2-\frac12=\frac32\)
=>\(x+\frac14=\frac94\)
=>x=2(nhận)
3 tháng 11 2021
Chú ý:
\(\left(x^2+2x\right)^2+4\left(x+1\right)^2=\left(x^2+2x\right)^2+4\left(x^2+2x+1\right)=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+4\)
\(=\left(x^2+2x+2\right)^2\)
\(x^2+\left(x+1\right)^2+\left(x^2+x\right)^2\)
\(=\left(x^2+x\right)+x^2+x^2+2x+1\)
\(=\left(x^2+x\right)^2+2x^2+2x+1\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)
\(=\left(x^2+x+1\right)^2\)
VJ
0
\(x^2\left(x^2+2\right)=4-x\sqrt{2x^2+4}\)
Đặt \(t=x\sqrt{2x^2+4}\)
Pttt: \(\dfrac{t^2}{2}=4-t\)
\(\Leftrightarrow t^2+2t-8=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\)
TH1: \(t=2\Rightarrow x\sqrt{2x^2+4}=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2\left(2x^2+4\right)=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^4+2x^2-2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x^2=-1+\sqrt{3}\end{matrix}\right.\)(do \(x^2\ge0\)) \(\Rightarrow x=\sqrt{-1+\sqrt{3}}\)
TH2: \(t=-4\Rightarrow x\sqrt{2x^2+4}=-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^2\left(2x^2+4\right)=16\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^4+2x^2-8=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le0\\x^2=2\end{matrix}\right.\)(do \(x^2\ge0\))\(\Rightarrow x=-\sqrt{2}\)
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