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b)đk:\(x\ge\dfrac{1}{2}\)
Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)
\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)
=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\)
Dấu = xảy ra\(\Leftrightarrow x=1\)
Vậy....
c) đk: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)
\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)
pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)
\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...
a)ĐKXĐ: x≥-1/3; x≤6
<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)
(vì x≥-1/3 nên3x+1≥0 )
\(\Leftrightarrow x^2+1-\left(x+3\right)\sqrt{x^2+1}+3x=0\)
Đặt \(\sqrt{x^2+1}=t>0\)
\(\Rightarrow t^2-\left(x+3\right)t+3x=0\)
\(\Delta=\left(x+3\right)^2-12x=\left(x-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{x+3+x-3}{2}=x\\t=\dfrac{x+3-x+3}{2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\left(x\ge0\right)\\\sqrt{x^2+1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\left(vô-nghiệm\right)\\x=\pm2\sqrt{2}\end{matrix}\right.\)
ĐK: Với mọi x thuộc R.
Ta có: \(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)
\(\Leftrightarrow\left(x^2+3x+1\right)^2=\left[\left(x+3\right)\sqrt{x^2+1}\right]^2\)
\(\Leftrightarrow x^4+6x^3+11x^2+6x+1=\left(x+3\right)^2\left(x^2+1\right)\)
\(\Leftrightarrow x^4+6x^3+11x^2+6x+1=x^4+6x^3+10x^2+6x+9\)
\(\Leftrightarrow x^2-8=0\)
\(\Leftrightarrow x^2=8\)
\(\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\end{matrix}\right.\)
Vậy....
ĐKXĐ: ...
\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)
\(\Leftrightarrow\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)-\sqrt{x+1}\left(\sqrt{3x-1}-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{3x-1}-\sqrt{x+1}\right)\left(\sqrt{3x-1}-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=\sqrt{x+1}\\\sqrt{3x-1}=x\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(a,PT\Leftrightarrow x^2-3x+2+x^2-x\sqrt{3x-2}=0\left(x\ge\dfrac{2}{3}\right)\\ \Leftrightarrow\left(x^2-3x+2\right)+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=0\\ \Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\)
Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)
Do đó \(x\in\left\{1;2\right\}\)
\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)
Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)
Vậy PT có nghiệm \(x=4\)
Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$
PT $\Leftrightarrow \frac{x}{\sqrt{x+2}}=\sqrt{3x+1}-\sqrt{x+1}$
$\Leftrightarrow \frac{x}{\sqrt{x+2}}=\frac{2x}{\sqrt{3x+1}+\sqrt{x+1}}$
$\Leftrightarrow x\left(\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}\right)=0$
Xét các TH:
TH1: $x=0$ (thỏa mãn)
TH2: $\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}$
$\Leftrightarrow \sqrt{3x+1}+\sqrt{x+1}=2\sqrt{x+2}$
$\Rightarrow 4x+2+2\sqrt{(3x+1)(x+1)}=4(x+2)$
$\Leftrightarrow \sqrt{(3x+1)(x+1)}=3$
$\Rightarrow (3x+1)(x+1)=9$
$\Leftrightarrow 3x^2+4x-8=0$
$\Rightarrow x=\frac{-2\pm 2\sqrt{7}}{3}$
Kết hợp với ĐKXĐ suy ra $x=\frac{-2+2\sqrt{7}}{3}$
Vậy............
a.
ĐKXĐ: \(x\ge2\)
\(\left(x+\sqrt{x}+1\right)\sqrt{x-2}=\left(x+1\right)^2-x\)
\(\Leftrightarrow\left(x+\sqrt{x}+1\right)\sqrt{x-2}=\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x-2}=x-\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x-2}+\sqrt{x}=x+1\)
\(\Leftrightarrow2x-2+2\sqrt{x^2-2x}=x^2+2x+1\)
\(\Leftrightarrow x^2-2\sqrt{x^2-2x}+3=0\)
\(\Leftrightarrow\left(\sqrt{x^2-2x}-1\right)^2+2x+2=0\) (vô nghiệm do \(2x+2>0\))
Vậy pt đã cho vô nghiệm
b. ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow2x^2-3x+1+2\left(x-1\right)\sqrt{2x^2-3x+1}+x^2-2x-3=0\)
Đặt \(\sqrt{2x^2-3x+1}=t\ge0\)
\(\Rightarrow t^2+2\left(x-1\right)t+x^2-2x-3=0\)
\(\Delta'=\left(x-1\right)^2-\left(x^2-2x-3\right)=4\)
\(\Rightarrow\left[{}\begin{matrix}t=1-x-2=-x-1\\t=1-x+2=3-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x^2-3x+1}=-x-1\left(x\le-1\right)\\\sqrt{2x^2-3x+1}=3-x\left(x\le3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=0\left(vn\right)\\x^2+3x-8=0\left(x\le3\right)\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{-3\pm\sqrt{41}}{2}\)