\(\Leftrightarrow30\sqrt{2}.\sqrt{x}=9\)

\(\Leftrightarrow\sqrt{x}=\dfrac{3\sqrt{2}}{20}\)

\(\Leftrightarrow x=\dfrac{9}{200}\)

Vậy pt có tập nghiệm là S=\(\left\{\dfrac{9}{200}\right\}\)

\(4\sqrt{2x}+21\sqrt{2x}=9-5\sqrt{2x}\)

\(30\sqrt{2x}=9\)

\(\sqrt{2x}=\dfrac{3}{10}\)

\(x=0.045\)

2√8x + 7√18x = 9 - √50x    Đk x >= 0

(=) 4√2x + 21√2x + 5√2x = 9

(=) 30 √2x = 9 

(=) √2x = 9/30

(=) 2x =9/100

(=) x = 9/200=0,045

Lời giải:

a) ĐK: $x\geq 2$

PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$

$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$

\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)

Vậy..........

b) ĐK: $x\geq 0$

PT $\Leftrightarrow (\sqrt{x}-3)^2=0$

$\Leftrightarrow \sqrt{x}-3=0$

$\Leftrightarrow x=9$ (thỏa mãn)

c) ĐK: $x\geq 3$

PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$

$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$

$\Leftrightarrow 3\sqrt{x-3}=7$

$\Leftrightarrow x-3=(\frac{7}{3})^2$

$\Rightarrow x=\frac{76}{9}$

d)

ĐK: $x\geq \frac{-1}{2}$

PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$

$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$

$\Leftrightarrow 3\sqrt{2x+1}=6$

$\Leftrightarrow \sqrt{2x+1}=2$

$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(3\sqrt{18x}-5\sqrt{8x}+4\sqrt{50x}=38\)

\(\Leftrightarrow9\sqrt{2x}-10\sqrt{2x}+20\sqrt{2x}=38\)

\(\Leftrightarrow19\sqrt{2x}=38\)

\(\Leftrightarrow\sqrt{2x}=2\)

\(\Leftrightarrow2x=4\)

hay x=2(thỏa ĐK)

b) ĐKXĐ: \(x\ge0\)

Ta có: \(3\sqrt{12x}-2\sqrt{27x}+4\sqrt{3x}=8\)

\(\Leftrightarrow6\sqrt{3x}-6\sqrt{3x}+4\sqrt{3x}=8\)

\(\Leftrightarrow\sqrt{3x}=2\)

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

c) ĐKXĐ: \(x\ge5\)

Ta có: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\)

hay x=9

a)

\(3.3\sqrt{2x}-5.2\sqrt{2x}+4.5.\sqrt{2x}=38\\ \Leftrightarrow19\sqrt{2x}=38\\ \Leftrightarrow\sqrt{2x}=2\\ \Leftrightarrow x=2\)

b)

\(3.2.\sqrt{3x}-2.3.\sqrt{3x}+4.\sqrt{3x}=8\\ \Leftrightarrow4\sqrt{3x}=8\\ \Leftrightarrow\sqrt{3x}=2\\\Leftrightarrow x=\dfrac{2^2}{3}=\dfrac{4}{3} \)

c)

\(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\)

`\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12`      `ĐK: x >= 1/2`

`<=>2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12`

`<=>4\sqrt{2x-1}=12`

`<=>\sqrt{2x-1}=3`

`<=>2x-1=9`

`<=>x=5` (t/m)

Vậy `S={5}`.

\(\Leftrightarrow2\sqrt{2x-1}-2\cdot3\sqrt{2x-1}+2\cdot4\sqrt{2x-1}=12\)

=>\(4\sqrt{2x-1}=12\)

=>\(\sqrt{2x-1}=3\)

=>2x-1=9

=>2x=10

=>x=5

\(\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12\) (ĐK: \(x\ge\dfrac{1}{2}\))

\(\Leftrightarrow2\sqrt{2x-1}-2\cdot3\sqrt{2x-1}+2\cdot4\sqrt{2x-1}=12\)

\(\Leftrightarrow2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12\)

\(\Leftrightarrow\left(2-6+8\right)\sqrt{2x-1}=12\)

\(\Leftrightarrow4\sqrt{2x-1}=12\)

\(\Leftrightarrow\sqrt{2x-1}=12:4\)

\(\Leftrightarrow\sqrt{2x-1}=3\)

\(\Leftrightarrow2x-1=9\)

\(\Leftrightarrow2x=9+1\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\left(tm\right)\)

Vậy \(x=5\)

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

http://123link.pro/eP1KS

9) Sửa: \(2\sqrt{8\sqrt{3}}-2\sqrt{5\text{ }\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=2\sqrt{2^2\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{2^2\cdot5\sqrt{3}}\)

\(=2\cdot2\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot2\sqrt{5\sqrt{3}}\)

\(=4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)

\(=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)

10) \(\sqrt{12x}-\sqrt{48x}-3\sqrt{3x}+27\)

\(=\sqrt{2^2\cdot3x}-\sqrt{4^2\cdot3x}-3\sqrt{3x}+27\)

\(=2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}+27\)

\(=-5\sqrt{3x}++27\)

11) \(\sqrt{18x}-5\sqrt{8x}+7\sqrt{18x}+28\)

\(=\sqrt{3^2\cdot2x}-5\sqrt{2^2\cdot2x}+7\sqrt{3^2\cdot2x}+28\)

\(=3\sqrt{2x}-5\cdot2\sqrt{2x}+7\cdot3\sqrt{2x}+28\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)

\(=14\sqrt{2x}+28\)

12) \(\sqrt{45a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{a}\)

\(=\sqrt{3^2\cdot5a}-\sqrt{2^2\cdot5a}+4\sqrt{3^2\cdot5a}+\sqrt{a}\)

\(=3\sqrt{5a}-2\sqrt{5a}+4\cdot3\sqrt{5a}+\sqrt{a}\)

\(=3\sqrt{5a}-2\sqrt{5a}+12\sqrt{5a}+\sqrt{a}\)

\(=13\sqrt{5a}+\sqrt{a}\)

\(\sqrt{2x-1}\) - \(\sqrt{8x-4}\) + \(\sqrt{50x-25}\) = 24 đk \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}\) - \(\sqrt{4.\left(2x-1\right)}\) + \(\sqrt{25.\left(2x-1\right)}\) = 24

\(\sqrt{2x-1}\) - 2\(\sqrt{2x-1}\) + 5\(\sqrt{2x-1}\) = 24

\(\sqrt{2x-1}\) (1 - 2 + 5) = 24

 4\(\sqrt{2x-1}\) = 24

   \(\sqrt{2x-1}\) = 24: 4

   \(\sqrt{2x-1}\) = 6

    \(2x-1=36\)

    2\(x\) = 37

      \(x=\dfrac{37}{2}\) (thỏa mãn)

Vậy \(x=\dfrac{37}{2}\)