x+ 3/4           = 5/6-7/12

x+ 3/4             = 1/4

x                      = 1/4-3/4

x                      = -1/2

5/6 - ( x + 3/4 ) = 7/12

       x + 3/4       = 5/6 - 7/12 

       x + 3/4       = 1/4

             x          = 1/4 - 3/4 

             x          = -1/2

\(\left|x-1\right|+\left|x-4\right|=3x\)

Trường hợp 1: Với x > 1

\(1-x+4-x=3x\)

\(\Leftrightarrow5-2x=3x\)

\(\Leftrightarrow5=3x+2x\)

\(\Leftrightarrow5=5x\)

\(\Leftrightarrow x=1\) (Loại)

Trường hợp 2: Với \(1\le x\le4\)

\(x-1+4-x=3x\)

\(\Leftrightarrow3=3x\)

\(\Leftrightarrow x=1\) (Thỏa mãn)

Trường hợp 3: Với x > 4

\(x-1+x-4=3x\)

\(\Leftrightarrow2x+5=3x\)

\(\Leftrightarrow2x-3x=5\)

\(\Leftrightarrow-x=5\)

\(\Leftrightarrow x=\left(-5\right)\) (Loại)

\(\frac{3}{4}\left(x-\frac{1}{2}\right)-\frac{5}{6}=\frac{1}{3}\)

\(\Leftrightarrow\frac{3}{4}x=\frac{3}{8}+\frac{5}{6}+\frac{1}{3}\)

\(\Leftrightarrow\frac{3}{4}x=\frac{37}{24}\)

\(\Leftrightarrow x=\frac{37}{18}\)

3/4.( x-1/2)     =  1/3+ 5/6

3/4.(x- 1/2)      = 7/6

 x- 1/2           = 7/6: 3/4

x- 1/2             = 14/9

x                     = 14/9+ 1/2

x                     = 37/18

\(A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{120}}\)

\(3A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{119}}\)

\(3A-A=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{119}}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{120}}\right)\)

\(=\frac{1}{3}-\frac{1}{3^{120}}\)

\(\Rightarrow A=\frac{1}{6}-\frac{1}{2.3^{120}}\)

\(\frac{a}{b}=\frac{c}{d}\)(\(b,d\ne0\))

\(\Leftrightarrow ad=bc\)

\(\Leftrightarrow2ad=2bc\)

\(\Leftrightarrow ad-bc=bc-ad\)

\(\Leftrightarrow ad-bc+ac-bd=bc-ad+ac-bd\)

\(\Leftrightarrow\left(a+b\right)\left(c-d\right)=\left(c+d\right)\left(a-b\right)\)

\(\Leftrightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(\(a-b,c-d\ne0\))