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: \(x\ge-1\))

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

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

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

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

\(\Leftrightarrow x+1=16\)

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

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) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)

ĐK : x ≥ 0

⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)

⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)

⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)

⇔ \(-3\sqrt{x}=-5\)

⇔ \(\sqrt{x}=15\)

⇔ \(x=225\)( tm )

b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)

ĐK : x ≤ 3

⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)

⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)

⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)

⇔ \(3\sqrt{3-x}=6\)

⇔ \(\sqrt{3-x}=2\)

⇔ \(3-x=4\)

⇔ \(x=-1\)( tm )

c) \(\sqrt{9x^2+12x+4}=4\)

⇔ \(\sqrt{\left(3x+2\right)^2}=4\)

⇔ \(\left|3x+2\right|=4\)

⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)

d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)

ĐK : x ≥ 1

⇔  \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)

⇔  \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)

⇔  \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)

⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)

⇔ \(\sqrt{x-1}=1\)

⇔ \(x-1=1\)

⇔ \(x=2\)( tm )

ĐK: \(x\ge0\)\(4\sqrt{x}-2\sqrt{9x}+16\sqrt{x}=5\)  5  (=) \(\sqrt{x}\left(4-2\sqrt{9}+16\right)=5\) (=) \(\sqrt{x}.14=5\)(=) x=\(\frac{25}{196}\)

ĐK: \(x\ge-5\)PT(=) \(\sqrt{5+x}\left(\sqrt{4}-3+\frac{4}{3}.3\right)=6\) (=) \(\sqrt{5+x}.3=6\) (=)\(\sqrt{5+x}=2\)(=) X = -1 (nhận)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

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

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a) \(\sqrt{\left(2x-1\right)^2}=3\)

⇔ \(\left|2x-1\right|=3\)

⇔ \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

⇔ \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

b) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)

ĐKXĐ : \(x\ge0\)

⇔ \(3\sqrt{x}-2\sqrt{3^2x}+\sqrt{4^2x}=5\)

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

⇔ \(7\sqrt{x}-6\sqrt{x}=5\)

⇔ \(\sqrt{x}=5\)

⇔ \(x=25\)( tm )

c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)

ĐKXĐ : \(x\ge-5\)

⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)

⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot3\sqrt{x+5}=6\)

⇔ \(-\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)

⇔ \(\frac{5}{4}\sqrt{x+5}=6\)

⇔ \(\sqrt{x+5}=\frac{24}{5}\)

⇔ \(x+5=\frac{576}{25}\)

⇔ \(x=\frac{451}{25}\left(tm\right)\)

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

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

\(\Leftrightarrow x=1\)