1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

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

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

\(\frac{5}{3}-\left(2x-\frac{2}{4}\right)\ge x-\left(4x-\frac{3}{6}\right)\)

\(\Leftrightarrow\frac{5}{3}-2x+\frac{1}{2}\ge x-4x+\frac{1}{2}\)

\(\Leftrightarrow x\ge-\frac{5}{3}\)

Ý c cx vậy nha ! Chuyển vế rồi thu gọn lại 

a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)

\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)

\(\Leftrightarrow-297-99x=0\)

\(\Leftrightarrow x=3\)

Vậy \(n_0\) của PT là: x=3

b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)

\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)

\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)

\(\Leftrightarrow-64-36x=250-30x\)

\(\Leftrightarrow-6x=314\)

\(\Leftrightarrow x=-\frac{157}{3}\)

Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)

c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)

\(\Leftrightarrow-3x=4x-\frac{23}{5}\)

\(\Leftrightarrow7x=\frac{23}{5}\)

\(\Leftrightarrow x=\frac{23}{35}\)

Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)

d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow x=-\frac{5}{12}\)

Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)

a) \(3\left(x-3\right)-5\left(-x+1\right)=x+6\)

\(\Leftrightarrow3x-9+5x-5-x-6=0\)

\(\Leftrightarrow7x=20\)

\(\Rightarrow x=\frac{20}{7}\)

b) \(\left|4x-2\right|=8\Leftrightarrow\orbr{\begin{cases}4x-2=8\\4x-2=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=10\\4x=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

c) \(-3\left|6x+1\right|=-12\)

\(\Leftrightarrow\left|6x+1\right|=4\Leftrightarrow\orbr{\begin{cases}6x+1=4\\6x+1=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}6x=3\\6x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{6}\end{cases}}\)

                                                   Bài giải

a, \(3\left(x-3\right)-5\left(-x+1\right)=x+6\)

\(3x-9+5x-5-x-6=0\)

\(7x-20=0\)

\(7x=20\)

\(x=\frac{20}{7}\)

b, \(\left|4x-2\right|=8\)

\(4x-2=\pm8\)

\(\Rightarrow\orbr{\begin{cases}4x-2=-8\\4x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-6\\4x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy \(x\in\left\{-3\text{ ; }2\right\}\)

c, \(-3\left|6x+1\right|=-12\)

\(\left|6x+1\right|=4\)

\(6x+1=\pm4\)

\(\Rightarrow\orbr{\begin{cases}6x+1=-4\\6x+1=4\end{cases}}\Rightarrow\orbr{\begin{cases}6x=-5\\6x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{6}\\x=\frac{1}{2}\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{-\frac{5}{6}\text{ ; }\frac{1}{2}\right\}\)

\(ĐK:x\ge2\\ PT\Leftrightarrow2\sqrt{x-2}+\sqrt{x-2}=6\\ \Leftrightarrow3\sqrt{x-2}=6\\ \Leftrightarrow\sqrt{x-2}=2\\ \Leftrightarrow x-2=4\\ \Leftrightarrow x=6\left(tm\right)\)

c: =>(x+2)(x+3)(x-5)(x-6)=180

=>(x^2-3x-10)(x^2-3x-18)=180

=>(x^2-3x)^2-28(x^2-3x)=0

=>x(x-3)(x-7)(x+4)=0

=>\(x\in\left\{0;3;7;-4\right\}\)

c: =>(x-3)(x+2)(2x+1)(3x-1)=0

=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

x²+10x+25-4x(x+5)=0

⇔(x+5)²-4x(x+5)=0

⇔(x+5)(x+5-4x)=0

⇔(x+5)(5-3x)=0

⇔\(\left\{{}\begin{matrix}x+5=0\\5-3x=0\end{matrix}\right.\Leftrightarrow\left\{{} }\left\{{}\begin{matrix}x=-5\\x=\dfrac{5}{3}\end{matrix}\right.\)