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

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

\(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\cdot\sqrt{x-3}=20\)

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

\(\Leftrightarrow x-3=25\)

hay x=28

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

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

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

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

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

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

\(=\sqrt{\left(x^2-2.x.3+3^2\right)}\)

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

\(=\sqrt{\left(x^2-2.x.1+1^2\right)}\)

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

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: \(x\ge-1\))

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

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

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

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

\(\Leftrightarrow x+1=16\)

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

\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)

\(< =>2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)

\(< =>2\cdot3\sqrt{\left(x-3\right)}-\frac{1}{5}.5\sqrt{\left(x-3\right)}-\frac{1}{7}.7\sqrt{\left(x-3\right)}=20\) \(đk:x\ge0\)

\(< =>6\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}=20\)

\(< =>\sqrt{\left(x-3\right)}\left(6-1-1\right)=20\)

\(< =>4\sqrt{\left(x-3\right)}=20\)

\(< =>\sqrt{\left(x-3\right)}=5\)

\(< =>x-3=25\)

\(< =>x=28\left(tm\right)\)

a) \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\) (ĐKXĐ : \(x\ge2\) )

\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}-4\sqrt{x+3}-3\sqrt{x-2}=2\)

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

\(\Leftrightarrow x+3=4\)

\(\Leftrightarrow x=1\) ( Thỏa mãn ĐKXĐ )

c) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\) (ĐKXĐ : \(x\ge-5\) )

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

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

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

\(\Leftrightarrow x+5=4\)

\(\Leftrightarrow x=-1\) ( Thỏa mãn ĐKXĐ )

Vậy.......