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a: Ta có: \(\sqrt{x^2-x+3}+7=10\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2-4x+8}-7=-5\)
\(\Leftrightarrow x^2-4x+8=4\)
\(\Leftrightarrow x-2=0\)
hay x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
b. 2 + \(\sqrt{2x-1}=x\) ĐKXĐ: \(x\ge0,5\)
<=> \(\sqrt{2x-1}\) = x - 2
<=> 2x - 1 = (x - 2)2
<=> 2x - 1 = x2 - 4x + 4
<=> -x2 + 2x + 4x - 4 - 1 = 0
<=> -x2 + 6x - 5 = 0
<=> -x2 + 5x + x - 5 = 0
<=> -(-x2 + 5x + x - 5) = 0
<=> x2 - 5x - x + 5 = 0
<=> x(x - 5) - (x - 5) = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. ĐKXĐ: $x\geq 0$
$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$
$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$
$\Leftrightarrow 13\sqrt{2x}=28$
$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$
$\Leftrightarrow 2x=\frac{784}{169}$
$\Leftrightarrow x=\frac{392}{169}$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$
PT $\Leftrightarrow \frac{3x-2}{x+1}=9$
$\Rightarrow 3x-2=9(x+1)$
$\Leftrightarrow x=\frac{-11}{6}$ (tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: ...
\(\Leftrightarrow3\left(2\sqrt{x+2}+\sqrt{3-x}\right)=3x+1+4\sqrt{-x^2+x+6}\)
Đặt \(2\sqrt{x+2}+\sqrt{3-x}=t>0\)
\(\Rightarrow t^2=4\left(x+2\right)+3-x+4\sqrt{\left(x+2\right)\left(3-x\right)}=3x+11+4\sqrt{-x^2+x+6}\)
Pt trở thành:
\(3t=t^2-10\)
\(\Leftrightarrow t^2-3t-10=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2\sqrt{x+2}+\sqrt{3-x}=5\)
Ta có: \(VT=2\sqrt{x+2}+\sqrt{3-x}\le\sqrt{\left(2^2+1^2\right)\left(x+2+3-x\right)}=5\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\frac{\sqrt{x+2}}{2}=\sqrt{3-x}\Leftrightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
a) ĐKXĐ: \(x\ge2\)
\(pt\Leftrightarrow x-2=x^2+2x+1\)
\(\Leftrightarrow x^2+x+3=0\)(vô lý do \(x^2+x+3=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\))
Vậy \(S=\varnothing\)
b) ĐKXĐ: \(x\ge-3\)
\(pt\Leftrightarrow1+x^2=x^2+6x+9\)
\(\Leftrightarrow6x=-8\Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)