a) \(\sqrt{36\left(x-5\right)^2}\left(x\ge5\right)=6\left|x-5\right|=6\left(x-5\right)=6x-30\)

b) \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}\left(x>1\right)=\dfrac{1}{2}\left|1-x\right|=\dfrac{1}{2}\left(x-1\right)=\dfrac{1}{2}x-\dfrac{1}{2}\)

c) \(\sqrt{x^2\left(2x-4\right)^2}\left(x\ge2\right)=\left|x\left(2x-4\right)\right|=x\left(2x-4\right)=2x^2-4x\)

d) \(\dfrac{1}{x}\sqrt{x^2\left(1+x\right)^2}\left(x< -1\right)=\dfrac{1}{x}\left|x\left(1+x\right)\right|=\dfrac{1}{x}x\left(1+x\right)=1+x\)

Bài 3:
a) \(\sqrt{3x-2}=4\)
⇔\(\sqrt{3x-2}=\sqrt{4^2}\)
⇔\(3x-2=4^2=16\)
    \(3x=16+2=18\)
    \(x=18:3=6\)
    Vậy \(x=6\)
b)\(\sqrt{4x^2+4x+1}-11=5\)
⇔\(\sqrt{\left(2x\right)^2+2\left(2x\right)\cdot1+1^2}-11=5\)
⇔\(\sqrt{\left(2x+1\right)^2}-11=5\)
TH1:
⇔\(\left(2x+1\right)-11=5\)
    \(2x+1=5+11=16\)
    \(2x=16-1=15\)
    \(x=15:2=7,5\)
TH2:
⇔\(\left(2x+1\right)-11=-5\)
    \(2x-1=-5+11=6\)
    \(2x=6+1=7\)
    \(x=7:2=3,5\)
    Vậy \(x=\left\{7,5;3,5\right\}\) 
    (Câu này mình không chắc chắn lắm)   
    (Học sinh lớp 6 đang làm bài này)    

Bài 4:

a: \(C=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)

\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\sqrt{x}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)

b: C-6<0

=>C<6

=>\(2\sqrt{x}< 6\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)

bài 1: 

a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)

\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)

\(=-33\sqrt{2}\)

b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)

\(=10-2\sqrt{21}+14\sqrt{21}\)

\(=12\sqrt{21}+10\)

Bài 2: 

a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)

\(\Leftrightarrow\left|2x+3\right|=8\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)

b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)

\(\Leftrightarrow4\sqrt{x}=8\)

hay x=4

c: Ta có: \(\sqrt{9x-9}+1=13\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

\(\Leftrightarrow x-1=16\)

hay x=17

Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?

Câu 1:ĐK \(x\ge\frac{1}{2}\)

\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)

<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)

<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)

<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)

<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)

<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)

Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)

=> \(x=1\)(TM ĐKXĐ)

Vậy x=1

1:

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0\)

=>x-3=0 hoặc \(\sqrt{x+3}=2\)

=>x=3 hoặc x+3=4

=>x=1(loại) hoặc x=3(nhận)

2:

\(\Leftrightarrow\left(\sqrt{4x+1}-\sqrt{3x-4}\right)^2=1\)

=>\(4x-1+3x-4-2\sqrt{\left(4x+1\right)\left(3x-4\right)}=1\)

=>\(\sqrt{4\left(4x+1\right)\left(3x-4\right)}=7x-6\)

=>4(12x^2-16x+3x-4)=(7x-6)^2

=>49x^2-84x+36=48x^2-52x-16

=>-84x+36=-52x-16

=>-32x=-52

=>x=13/8

3: =>\(\sqrt{\left(x-5\right)^2}=5-x\)

=>|x-5|=5-x

=>x-5<=0

=>x<=5

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

=>\(\left\{{}\begin{matrix}x>=-2\\\left(x-4\right)^2=\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\x^2-8x+16=x^2+4x+4\end{matrix}\right.\)

=>x>=-2 và -8x+16=4x+4

=>x=1

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

=>\(\sqrt{x-2}=10\)

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

\(\sqrt{4\left(x-3\right)^2}=8\)

=>\(\sqrt{\left(2x-6\right)^2}=8\)

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

=>\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

a) ĐKXĐ: \(x\ge0\)

Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)

\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)

Dòng thứ 2 qua dòng thứ 3 anh làm chậm lại được không ạ, tại tắt quá e không hiểu