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a) \(\sqrt{1-6x+9x^2}=9\)
\(\Leftrightarrow\sqrt{\left(1-3x\right)^2}=9\)
\(\Leftrightarrow\left|1-3x\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3x=9\\1-3x=-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=1-9\\3x=1+9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-8\\3x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=\dfrac{10}{3}\end{matrix}\right.\)
b) \(\sqrt{2x-3}-\sqrt{x+1}=0\) (\(x\ge\dfrac{3}{2}\))
\(\Leftrightarrow\sqrt{2x-3}=\sqrt{x+1}\)
\(\Leftrightarrow2x-3=x+1\)
\(\Leftrightarrow2x-x=1+3\)
\(\Leftrightarrow x=4\left(tm\right)\)
c) \(\sqrt{9x^2+12+4}-2=3x\)
\(\Leftrightarrow\sqrt{\left(3x+2\right)^2}=3x+2\)
\(\Leftrightarrow\left|3x+2\right|=3x+2\)
\(\Leftrightarrow3x+2\ge0\)
\(\Leftrightarrow3x\ge-2\)
\(\Leftrightarrow x\ge-\dfrac{2}{3}\)
a: =>|3x-1|=9
=>3x-1=9 hoặc 3x-1=-9
=>x=-8/3 hoặc x=10/3
b: =>căn 2x-3=căn x+1
=>2x-3=x+1
=>x=4
c: =>|3x+2|=3x+2
=>3x+2>=0
=>x>=-2/3
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
\(a,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{3x+4}=\sqrt{2x+1}+1\\ \Leftrightarrow3x+4=2x+2+2\sqrt{2x+1}\\ \Leftrightarrow x+2=2\sqrt{2x+1}\\ \Leftrightarrow x^2+4x+4=8x+4\\ \Leftrightarrow x^2-4x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\\ b,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{2x-1}=2\sqrt{x-1}-1\\ \Leftrightarrow2x-1=4x-3-4\sqrt{x-1}\\ \Leftrightarrow2x-2-4\sqrt{x-1}=0\\ \Leftrightarrow x-1-2\sqrt{x-1}=0\\ \Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
a, ĐK: \(x\ge0\)
\(\sqrt{2x}-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=25\left(tm\right)\)
a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
a. ĐKXĐ: $x\geq 2$ hoặc $x=1$
PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$
$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)
b.
PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$
$\Leftrightarrow |x-2|=|2x-3|$
\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)
c. ĐKXĐ: $x=2$ hoặc $x\geq 3$
PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)
d.
PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$
$\Leftrightarrow |2x-1|=|x-3|$
\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)
a.
ĐKXĐ: \(x\le\dfrac{2}{3}\)
\(3x^2-7x+2-\left(1-\sqrt{2-3x}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)-\dfrac{3x-1}{1+\sqrt{2-3x}}=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-2-\dfrac{1}{1+\sqrt{2x-3}}\right)=0\) (1)
Do \(x\le\dfrac{2}{3}\Rightarrow x-2< 0\Rightarrow x-2-\dfrac{1}{1+\sqrt{2-3x}}< 0;\forall x\in TXĐ\)
Nên (1) tương đương:
\(3x-1=0\Leftrightarrow x=\dfrac{1}{3}\)
b.
ĐKXĐ: \(x\ge-\dfrac{1}{2}\)
\(18x^2+6x+3=9x\sqrt{6x+3}\)
Đặt \(\sqrt{6x+3}=y\ge0\) ta được:
\(18x^2+y^2=9xy\)
\(\Leftrightarrow18x^2-9xy+y^2=0\)
\(\Leftrightarrow\left(6x-y\right)\left(3x-y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=3x\\y=6x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{6x+3}=3x\\\sqrt{6x+3}=6x\end{matrix}\right.\) (\(x\ge0\))
\(\Leftrightarrow\left[{}\begin{matrix}6x+3=9x^2\\6x+3=36x^2\end{matrix}\right.\) (\(x\ge0\))
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1+\sqrt{13}}{12}\end{matrix}\right.\)