\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

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

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

a: ĐKXĐ: x-5>=0

=>x>=5

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

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x-1>=0

=>x>=1

\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)

=>\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)

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

=>\(\sqrt{x-1}=-2\)(vô lý)

Vậy: Phương trình vô nghiệm

c: ĐKXĐ: x-2>=0

=>x>=2

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)

=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)

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

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

=>x-2=16

=>x=18(nhận)

d: ĐKXĐ: x+3>=0

=>x>=-3

\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)

=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)

=>\(4\sqrt{x+3}=0\)

=>x+3=0

=>x=-3(nhận)

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

= \(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)

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

= \(2\sqrt{x-5}=4\)

= \(\sqrt{x-5}=2\)

= \(\left|x-5\right|=4\)

=> \(x-5=\pm4\)

\(x=\pm4+5\)

\(x=9;x=1\)

Vậy x=9; x=1

b) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)

\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)

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

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

=>\(\left|x-1\right|=-2\)

\(x-1=\mp2\)

\(x=-3;x=1\)

Vậy x=-3; x=1

a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

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

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

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

=>\(\dfrac{3}{2}\sqrt{x-5}=3\)

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

=>x-5=4

=>x=9(nhận)

d. \(\sqrt{9x^2+12x+4}=4\)

<=> \(\sqrt{\left(3x+2\right)^2}=4\)

<=> \(|3x+2|=4\)

<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)

\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)

\(\Leftrightarrow x=1\)

a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

=>\(\sqrt{\left(x-2\right)^2}=7\)

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

=>\(3\sqrt{x+3}=6\)

=>\(\sqrt{x+3}=2\)

=>x+3=4

=>x=1(nhận)

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrtx{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrt{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

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

<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7