1. ĐKXĐ: \(x\ge3\)

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

⇔ \(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

⇔ \(4\sqrt{x-3}=20\)

⇔ \(\sqrt{x-3}=5\)

⇔ \(x-3=25\)

⇔ \(x=28\left(TMĐKXĐ\right)\)

Vậy....

2. ĐKXĐ: \(x\ge0\)

\(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-7=\frac{1}{2}\sqrt{5x}\)

⇔ \(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-\frac{1}{2}\sqrt{5x}=7\)

⇔ \(2\sqrt{5x}=7\)

⇔ \(\sqrt{5x}=\frac{7}{2}\)

⇔ \(5x=\frac{49}{4}\)

⇔ \(x=\frac{49}{20}\left(TMĐKXĐ\right)\)

Vậy...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow x-3=25\)

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: 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) 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)

1 slot xíu nữa làm :))))) 

8h lên giúp bạn trước rồi giúp mấy bạn khác sau :v

a, nhóm can x vào một nhóm cái trong ngoặc còn lại thì tính ra

\(11\sqrt{5x}=33\)

chia cả hai vế cho 11 căn 5 rồi bình phương hai vế do x>=0

b,sai đề

Làm 1 câu hay 2 câu nhờ :)))))) Tương tự 1 câu câu còn lại tự làm lấy nhé ;v

\(3\sqrt{9x-27}+2\sqrt{25x-75}\)\(-\sqrt{49x-147}=48\)

\(\Leftrightarrow3\sqrt{9\left(x-3\right)}+2\sqrt{25\left(x-3\right)}\)\(-\sqrt{49\left(x-3\right)}=48\)

\(\Leftrightarrow9\sqrt{x-3}+10\sqrt{x-3}-7\sqrt{x-3}\)\(=48\)

\(\Leftrightarrow\sqrt{x-3}=48\)

\(\Leftrightarrow x-3=16\Leftrightarrow x=19\)

P/s câu trên cx tương tự như trên :)))))) Nhưng có điều hơi khác 1 xíu