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ĐKXĐ: \(x\ge-\frac{1}{3}\)
\(\Leftrightarrow\left(\sqrt{3x+1}\right)^3=\left(x^2+1\right)^3\)
\(\Leftrightarrow\sqrt{3x+1}=x^2+1\)
\(\Leftrightarrow3x+1=x^4+2x^2+1\)
\(\Leftrightarrow x^4+2x^2-3x=0\)
\(\Leftrightarrow x\left(x^3+2x-3\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
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a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\x^2+6x+9=21-x^2-4x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\2x^2+10x-12=0\end{matrix}\right.\Leftrightarrow x=1\)
b: \(\left|x^2+5x+4\right|-4=x\)
=>|x^2+5x+4|=x+4
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-4\\\left(x^2+5x+4-x-4\right)\left(x^2+5x+4+x+4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-4\\\left(x^2+4x\right)\left(x^2+6x+8\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;-2;-4\right\}\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x^2-5x+4-2x+1\right)\left(2x^2-5x+4+2x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x^2-7x+5\right)\left(2x^2-3x+3\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{5}{2};1\right\}\)
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Lời giải:ĐK: $x> \frac{1}{3}$ hoặc $x<0$
Đặt $\sqrt{\frac{3x-1}{x}}=a(a> 0)$ thì BPT trở thành:
$2a\geq \frac{1}{a^2}+1$
$\Leftrightarrow 2a^3\geq a^2+1$
$\Leftrightarrow (a-1)(2a^2+a+1)\geq 0$
$\Leftrightarrow a\geq 1$
$\Leftrightarrow \sqrt{\frac{3x-1}{x}}\geq 1$
$\Leftrightarrow \frac{3x-1}{x}\geq 1(*)$
Nếu $x>\frac{1}{3}$ thì $(*)\Leftrightarrow 3x-1\geq x\Leftrightarrow x\geq \frac{1}{2}$
Nếu $x< 0$ thì $(*)\Leftrightarrow 3x-1\leq x\Leftrightarrow x\leq \frac{1}{2}\Rightarrow x< 0$
Vậy $x\geq \frac{1}{2}$ hoặc $x< 0$
![](https://rs.olm.vn/images/avt/0.png?1311)
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Đặt \(\sqrt{1-3x}=a;\sqrt{x^2+1}=b\left(b>0;a\ge0\right)\)
\(\sqrt{2x^2+3x+1}=\sqrt{2\left(x^2+1\right)+\left(3x-1\right)}=\sqrt{2b^2-a^2}\)
\(\Leftrightarrow\sqrt{2b^2-a^2}+a=2b\)
\(\Leftrightarrow\sqrt{2b^2-a^2}=2b-a\) (2b ≥ a)
Bình phương lên:
\(2b^2-a^2=4b^2-4ab+a^2\)
\(\Leftrightarrow2b^2+2a^2-4ab=0\)
\(\Leftrightarrow a^2+b^2-2ab=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)
Tự giải tiếp đc ko ạ ??
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ĐK:\(-1\le x\le\dfrac{1}{3}\)
Ta có: VT=\(\sqrt{2x^2+3x+1}+\sqrt{1-3x}\le\sqrt{\left(1+1\right)\left(\sqrt{2x^2+3x+1}^2+\sqrt{1-3x}^2\right)}\)
\(=\sqrt{2.\left(2x^2+2\right)}=2\sqrt{x^2+1}\)
Xét VT= \(2\sqrt{2x^2+1}\ge2\sqrt{x^2+1}\)
\(\Leftrightarrow2x^2+1\ge x^2+1\Leftrightarrow x^2\ge0\) (đúng)
\(\Rightarrow VP\ge VT\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x^2+3x+1}}=\dfrac{1}{\sqrt{1-3x}}\\2\sqrt{2x^2+1}=2\sqrt{x^2+1}\end{matrix}\right.\Leftrightarrow x=0\left(tm\right)\)