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

pt <=> \(\sqrt{4.\left(x-5\right)^2}=0\)

=> \(2.lx-5l=0\)

=> \(lx-5l=0\)

=> x - 5 = 0

=> x = 5 

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

\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+16}=\dfrac{\left(x+16\right)\left(\sqrt{x}+2\right)}{\left(x-16\right)\left(\sqrt{x}+16\right)}\)

ĐK: x \(\ge\) 1

Có:

\(=\sqrt{4}.\sqrt{x-1}-\sqrt{9}.\sqrt{x-1}-\sqrt{16}.\sqrt{x-1}\\ =2.\sqrt{x-1}-3.\sqrt{x-1}-4.\sqrt{x-1}\\ =\sqrt{x-1}\left(2-3-4\right)\\ =-5\sqrt{x-1}\)

\(=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}\)

\(=-5\sqrt{x-1}\)

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

\(\Leftrightarrow\left|x-2\right|=\sqrt{x-2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{x-2}\\-x+2=\sqrt{x-2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

Vậy ....

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