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)

`sqrt{x-5}+2sqrt{4x-20}-1/2sqrt{9x-45}=12`

Điều kiện:`x>=5`

`pt<=>sqrt{x-5}+2sqrt{4(x-5)}-1/2sqrt{9(x-5)}=12`

`<=>sqrt{x-5}+4sqrt{x-5}-3/2sqrt{x-5}=12`

`<=>7/2sqrt{x-5}=12`

`<=>sqrt{x-5}=24/7`

`<=>x-5=576/49`

`<=>x=821/49(Tmđk)`

Vậy `S={821/49}.`

Ta có: \(\sqrt{x-5}+2\sqrt{4x-20}-\dfrac{1}{3}\sqrt{9x-45}=12\)

\(\Leftrightarrow4\sqrt{x-5}=12\)

\(\Leftrightarrow x-5=9\)

hay x=14

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: \(x\ge0\)

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)

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

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 : \(x\ge-5\)

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

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

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

⇔ \(\frac{5}{4}\sqrt{x+5}=6\)

⇔ \(\sqrt{x+5}=\frac{24}{5}\)

⇔ \(x+5=\frac{576}{25}\)

⇔ \(x=\frac{451}{25}\left(tm\right)\)

Giải:

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

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

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

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

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

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

\(\Leftrightarrow\)\(\sqrt{x-5}\)= 2 . Đk : x \(\ge\)5

\(\Rightarrow\)x - 5 = 4

\(\Leftrightarrow\)x = 9 ( thỏa mãn )

Vậy phương trình đã cho có tập nghiệm S = \(\left\{9\right\}\)

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)