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)

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) 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

a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)

\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)

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

\(\Leftrightarrow x+3=9\)

hay x=6

b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)

\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;-1\right\}\)

\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)

\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)

Vậy \(S=\left\{28\right\}\)

a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)

\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)

\(\Leftrightarrow25x-4x=-8-75\)

\(\Leftrightarrow21x=-83\)

hay \(x=-\dfrac{83}{21}\)

b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)

\(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)

\(\Leftrightarrow\left|2x+1\right|=3x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)

d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)

\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)

\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)

\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)

\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)

\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)

\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)

vậy: Phương trình vô nghiệm

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

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

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

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

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

\(ĐK:x\ge5\)

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

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

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

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

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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))

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

c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)

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

\(\Leftrightarrow x-1=4\)

hay x=5

e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)

\(\Leftrightarrow\left|2x-7\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)

a. ĐKXĐ: $x\in\mathbb{R}$

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

$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$

$\Leftrightarrow x\leq 2$

b. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$

$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$

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

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

$\Leftrightarrow \frac{1}{4}=x-2$

$\Leftrightarrow x=\frac{9}{4}$ (tm)

c. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4$

$\Leftrightarrow 2\sqrt{x-1}=4$

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

$\Leftrightarrow x-1=4$

$\Leftrightarrow x=5$ (tm)

d. ĐKXĐ: $x\geq 2$

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

$\Leftrightarrow \frac{1}{2}\sqrt{x-2}-\frac{8}{3}\sqrt{x-2}+3\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{5}{6}\sqrt{x-2}-5=0$

$\Leftrightarrow \sqrt{x-2}=6$

$\Leftrightarrow x-2=36$

$\Leftrightarrow x=38$ (tm)

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: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)