a: ĐKXĐ: \(\left\{{}\begin{matrix}x+6>=0\\x-2>=0\end{matrix}\right.\Leftrightarrow x>=2\)

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

=>\(\left(\sqrt{x+6}-\sqrt{x-2}\right)^2=4\)

=>\(x+6+x-2-2\sqrt{\left(x+6\right)\left(x-2\right)}=4\)

=>\(2\sqrt{\left(x+6\right)\left(x-2\right)}=2x+4-4=2x\)

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

=>\(\left\{{}\begin{matrix}x>=0\\\left(x+6\right)\left(x-2\right)=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=2\\x^2+4x-12=x^2\end{matrix}\right.\)

=>x=3

b: ĐKXĐ: \(x-3>=0\)

=>x>=3

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

=>\(\sqrt{4x-12}=2x-3\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\4x-12=4x^2-12x+9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\4x^2-12x+9-4x+12=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\4x^2-16x+21=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

\(a,\left|2x+2\right|+10=2x\)

*TH1 : \(\left|2x+2\right|=2x+2\Leftrightarrow2x+2>0\Leftrightarrow x>-1\)

\(\Rightarrow2x+2+10=2x\)

\(\Leftrightarrow2x-2x=-10-2\)

\(\Leftrightarrow0x=-12\left(vô\cdot lý\right)\)

*TH2 :\(\left|2x+2\right|=-2x-2\Leftrightarrow-2x-2< 0\Leftrightarrow x>-1\)

\(\Rightarrow-2x-2+10=2x\)

\(\Leftrightarrow-2x-2x=-10+2\)

\(\Leftrightarrow-4x=-8\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(nhận\right)\)

Vậy \(S=\left\{\dfrac{1}{2}\right\}\)

\(b,\left|x-6\right|=\left|3-2x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=3-2x\\x-6=-3+2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{-3;3\right\}\)

a/

\(\Leftrightarrow x^2-2x+4-4=0\\ \Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow x=0;x-2=0\)

\(\Leftrightarrow x=0;x=2\)

b/

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

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

\(\Leftrightarrow\left(x-3\right)\left(3-x\right)=0\)

\(\Rightarrow x=3\)

\(a,x^2-2.\left(x-2\right)=4\\ \Leftrightarrow x^2-2x+4-4=0\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ \Rightarrow S=\left\{0;2\right\}\\ b,x^2-9-2x\left(x-3\right)=0\\ \Leftrightarrow x^2-4x^2+6x-9=0\\ \Leftrightarrow-3x^2+6x-9=0\\ \Leftrightarrow x^2-2x+3=0\\ \Leftrightarrow\left(x^2-2x+1\right)+2=0\\ \Leftrightarrow\left(x-1\right)^2=-2\left(vô.lí\right)\\ \Rightarrow Pt.vô.nghiệm\)

\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)

\(a,\left(đk:x\ge0\right)\) 

\(x=0\Rightarrow\sqrt{0+3}+0=0\left(vô-nghiệm\right)\)

\(x>0\)

\(\)\(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}+\dfrac{4\sqrt{x}}{\sqrt{x+3}}=4\)

\(VT\ge2\sqrt{\dfrac{\sqrt{x+3}}{\sqrt{x}}.\dfrac{4\sqrt{x}}{\sqrt{x+3}}}=4\)

\(dấu"="xảy-ra\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}=\dfrac{4\sqrt{x}}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Leftrightarrow x=1\left(tm\right)\)

\(b.2x^4-5x^3+6x^2-5x+2=0\Leftrightarrow\left(x-1\right)^2\left(2x^2-2x+2\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2-2x+2=0\left(vô-nghiệm\right)\end{matrix}\right.\)

a) ĐKXĐ : \(x\ge0\)

PT <=> \(x+3-4\sqrt{x}\sqrt{x+3}+4x=0\)

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

<=> \(\sqrt{x+3}=2\sqrt{x}\)

<=> \(x+3=4x\)

<=> x = 1

Vậy x = 1 là nghiệm phương trình

a: \(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)

=>x=0

b: \(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{x+1}{x-3}=\dfrac{1}{2}\)

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

\(\Leftrightarrow x^2-4x+3=2x^2-6x-2x^2+2=-6x+2\)

\(\Leftrightarrow x^2+2x+1=0\)

=>x=-1(nhận)

\(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)

\(\Leftrightarrow x=0\) ( vì \(x^2+2x+5>0;\forall x\)

b.\(\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\)

\(ĐK:x\ne1;3;4\)

\(\Leftrightarrow\dfrac{x}{\left(x-1\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\)

\(\Leftrightarrow\dfrac{x\left(x-3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-1\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)\)

\(\Leftrightarrow x^2-3x-\left(x^2-3x-x+3\right)=x^2-1\)

\(\Leftrightarrow x^2-3x-x^2+4x-3=x^2-1\)

\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow x^2-x+2x-2=0\)

\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{-2\right\}\)

\(a,2x^3+4x^2+10x=0\\ \Leftrightarrow2x\left(x^2+2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x^2+2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x^2+2x+1\right)+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2+4=0\left(vô..lí\right)\end{matrix}\right.\)

\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne3\\x\ne4\end{matrix}\right.\\ \dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x}{x-1}-\dfrac{1}{2}-\dfrac{x+1}{x-3}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2x^2-2}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-6x-x^2+4x-3-2x^2+2}{2\left(x-1\right)\left(x-3\right)}=0\)

\(\Rightarrow-x^2-2x-1=0\)

\(\Leftrightarrow x^2+2x+1=0\\ \Leftrightarrow\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\left(tm\right)\)

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)