8 - ( 6 + x ) = 3

8 - 6 + x      = 3

x + 8 - 6     = 3

x + 8          = 3 + 6  

x + 8         =  9

x                = 9 - 8

x                 = 1

mk nghĩ nên có 2 TH

8 - | 6 + x | = 3

=> | 6 + x | = 8 - 3

=> | 6 + x | = 5

=> 6 + x = 5 hoac 6 + x = - 5

=> x = 5 - 6 hoac x = - 5 - 6

=> x = - 1 hoac x = - 11

Vay x = - 1 , - 11

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

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

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

\(6x+1=\pm4\)

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

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

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

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

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

Vậy: S={-3}

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

chúc bạn học tốt

=>x-6=3-2x hoặc x-6=2x-3

=>3x=9 hoặc -x=3

=>x=3 hoặc x=-3

\(\left|x-6\right|=\left|3-2x\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-6=3-2x\\x-6=2x-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-6-3+2x=0\\2x-3-x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x-9=0\\x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)