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![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge0\)
\(\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
\(\Leftrightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
ĐK: \(x\in\mathbb{R}\)
\(\sqrt{x^2-2x+5}=x^2-2x-1=x^2-2x+5-6\)
Đặt \(\sqrt{x^2-2x+5}=t(t\geq 0)\). PT trở thành:
\(t=t^2-6\)
\(\Leftrightarrow t^2-t-6=0\Leftrightarrow (t-3)(t+2)=0\)
\(\Leftrightarrow \left[\begin{matrix} t=3\\ t=-2\end{matrix}\right.\). Vì $t\geq 0$ nên $t=3$
Do đó: \(\sqrt{x^2-2x+5}=3\Rightarrow x^2-2x+5=9\)
\(\Rightarrow x^2-2x-4=0\Rightarrow x=1\pm \sqrt{5}\)
Vậy........
Câu 2:
ĐK: \(x\in\mathbb{R}\)
Ta có: \(x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(\Rightarrow 2x^2-8x-12=2\sqrt{2x^2-8x+12}\)
\(\Leftrightarrow (2x^2-8x+12)-24-2\sqrt{2x^2-8x+12}=0\)
Đặt \(\sqrt{2x^2-8x+12}=t(t\geq 0)\). PT trở thành:
\(t^2-24-2t=0\)
\(\Leftrightarrow (t-6)(t+4)=0\Rightarrow \left[\begin{matrix} t=6\\ t=-4\end{matrix}\right.\)
Mà \(t\geq 0\Rightarrow t=6\)
Do đó: \(\sqrt{2x^2-8x+12}=6\Rightarrow 2x^2-8x+12=36\)
\(\Rightarrow x^2-4x-12=0\Rightarrow \left[\begin{matrix} x=6\\ x=-2\end{matrix}\right.\)
Vậy...........
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)
PT <=> 2x - 1 = 5
<=> x = 3 ( TM )
Vậy ...
b, ĐKXĐ : \(x\ge5\)
PT <=> x - 5 = 9
<=> x = 14 ( TM )
Vậy ...
c, PT <=> \(\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy ...
d, PT<=> \(\left|x-3\right|=3-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=x-3\\x-3=3-x\end{matrix}\right.\)
Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)
e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)
PT <=> 2x + 5 = 1 - x
<=> 3x = -4
<=> \(x=-\dfrac{4}{3}\left(TM\right)\)
Vậy ...
f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)
PT <=> \(x^2-x=3-x\)
\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )
Vậy ...
a) \(\sqrt{2x-1}=\sqrt{5}\) (x \(\ge\dfrac{1}{2}\))
<=> 2x - 1 = 5
<=> x = 3 (tmđk)
Vậy S = \(\left\{3\right\}\)
b) \(\sqrt{x-5}=3\) (x\(\ge5\))
<=> x - 5 = 9
<=> x = 4 (ko tmđk)
Vậy x \(\in\varnothing\)
c) \(\sqrt{4x^2+4x+1}=6\) (x \(\in R\))
<=> \(\sqrt{\left(2x+1\right)^2}=6\)
<=> |2x + 1| = 6
<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)
Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)