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\(1\))\(x^2+5x+8=3\sqrt{x^3+5x^2+7x+6}\left(1\right)\\ĐK:x\ge-\dfrac{3}{2} \\ \left(1\right)\Leftrightarrow x^2+5x+8=3\sqrt{\left(2x+3\right)\left(x^2+x+2\right)}\left(2\right)\)
Đặt \(b=\sqrt{2x+3};a=\sqrt{x^2+x+2}\)
\(\left(2\right)\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)\(\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1\pm\sqrt{5}}{2}\\x=\dfrac{7\pm\sqrt{89}}{2}\end{matrix}\right.\)
4)\(ĐK:x\ge-\dfrac{1}{3}\)
\(x^2-7x+2+2\sqrt{3x+1}=0\\ \Leftrightarrow x^2-7x+6+2\sqrt{3x+1}-4=0\\ \Leftrightarrow\left(x-1\right)\left(x-6\right)+\dfrac{12\left(x-1\right)}{2\sqrt{3x+1}+4}=0\\ \Leftrightarrow\left(x-1\right)\left(x-6+\dfrac{12}{2\sqrt{3x+1}+4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x-6+\dfrac{12}{2\sqrt{3x+1}+4}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x-5\right)+\dfrac{6}{\sqrt{3x+1}+2}-1=0\\ \Leftrightarrow\left(x-5\right)+\dfrac{4-\sqrt{3x+1}}{\sqrt{3x+1}+2}=0\\ \Leftrightarrow\left(x-5\right)-\dfrac{3\left(x-5\right)}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}=0\\ \Leftrightarrow\left(x-5\right)\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)=3\\ \Leftrightarrow3x+1+6\sqrt{3x+1}+8=3\\ \Leftrightarrow x+2\sqrt{3x+1}+2=0\\ \Leftrightarrow2\sqrt{3x+1}=-x-2\ge0\Leftrightarrow x\le-2\)
Vậy pt có 2 nghiệm là x=1 và x=5
b: ĐKXĐ: x>=-1
\(\sqrt{x+1}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x+1\right)^2=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\cdot x=0\\x>=-1\end{matrix}\right.\Leftrightarrow x\in\left\{0;-1\right\}\)
c: \(\sqrt{x-1}=1-x\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-1>=0\\1-x< =0\end{matrix}\right.\Leftrightarrow x=1\)
Do đó: x=1 là nghiệm của phương trình
d: \(2x+3+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\)(ĐKXĐ: x<>1)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)+4=x^2+3\)
\(\Leftrightarrow2x^2-2x+3x-3+4-x^2-3=0\)
\(\Leftrightarrow x^2+x-2=0\)
=>(x+2)(x-1)=0
=>x=-2(nhận) hoặc x=1(loại)
ĐKXĐ: \(x\ge-3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+2}=a>0\\\sqrt{x+3}=b\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x^2-x-5=2a^2-3b^2\\2x^2+x+1=2a^2-b^2\end{matrix}\right.\)
\(\Rightarrow\left(2a^2-3b^2\right)a+\left(2a^2-b^2\right)b\)
\(\Leftrightarrow2a^3+2a^2b-3ab^2-b^3=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a^2+4ab+b^2\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow x^2+x+2=x+3\Leftrightarrow x^2=1\)
Bài 1. ĐKXĐ:.........
PT \(\Leftrightarrow (-x^2+3x+3)+4\sqrt{-x^2+2x+3}=12\)
Đặt \(\sqrt{-x^2+2x+3}=t(t\geq 0)\) thì PT trở thành:
\(t^2+4t=12\)
\(\Leftrightarrow (t-2)(t+6)=0\Rightarrow \left[\begin{matrix} t=2\\ t=-6\end{matrix}\right.\)
Vì $t\geq 0$ nên $t=2$
$\Rightarrow -x^2+2x+3=t^2=4$
$\Leftrightarrow -x^2+2x-1=0$
$\Leftrightarrow -(x-1)^2=0\Leftrightarrow x=1$ (thỏa mãn)
Vậy......
Lời giải:
Ta thấy:
\(|x+2|\geq 0(1), \forall x\in\mathbb{R}\)
\(|x-2|+1\geq 1>0, \forall x\in\mathbb{R}\Rightarrow \frac{2}{|x-2|+1}>0(2)\)
Từ \((1);(2)\Rightarrow |x+2|+\frac{2}{|x-2|+1}>0\) với mọi $x\in\mathbb{R}$
Do đó PT đã cho vô nghiệm.
1) Phương trình đã cho tương đương
\(\Leftrightarrow\left(x-2\right)\left(3\sqrt{x^2+1}-x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\frac{3}{4}\end{matrix}\right.\)