Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

\(\frac{6}{11}\cdot\frac{3}{14}+\frac{-5}{16}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-3}{16}\)

\(=\frac{6}{11}\cdot\frac{3}{14}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-5}{16}+\frac{-3}{16}\)

\(=\frac{3}{14}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+\left(\frac{-5}{16}+\frac{-3}{16}\right)\)

\(=\frac{3}{14}\cdot\frac{6+5}{11}+\frac{-5+\left(-3\right)}{16}\)

\(=\frac{3}{14}\cdot\frac{11}{11}+\frac{-8}{16}\)

\(=\frac{3}{14}\cdot1+\frac{-1}{2}\)

\(=\frac{3}{14}+\frac{-1}{2}\)

\(=\frac{3}{14}+\frac{-7}{14}\)

\(=\frac{3+\left(-7\right)}{14}\)\(=\frac{-4}{14}=\frac{-2}{7}\)

\(\frac{-5}{6}+\left(7x-\frac{1}{2}\right)\cdot\frac{2}{9}=-1\frac{1}{3}\)

       \(\frac{-5}{6}+\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}\)

                       \(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}+\frac{5}{6}\)

                       \(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{1}{2}\)

                                   \(\frac{14}{9}\cdot x=-\frac{1}{2}+\frac{1}{9}\)

                                   \(\frac{14}{9}\cdot x=-\frac{7}{18}\)

                                              \(x=-\frac{7}{18}:\frac{14}{9}\)

                                              \(x=-\frac{1}{4}\)

a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)

\(x+\frac{2}{5}=\frac{-7}{20}\)

\(x=\frac{-13}{20}\)

Vậy \(x=\frac{-13}{20}\)

b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)

  \(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x=\frac{15}{8}\)

\(x=\frac{-15}{4}\)

Vậy \(x=\frac{-15}{4}\)

\(a,2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{11}{20}-\frac{5}{4}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)

\(x+\frac{2}{5}=\frac{-7}{10}:2\)

\(x+\frac{2}{5}=\frac{-7}{20}\)

\(x=\frac{-7}{20}-\frac{2}{5}\)

\(x=\frac{-3}{4}\)

b) (x+1)^3-x(x-2)^2+x-1=0

 ⇔x^3+3x^2+3x+1-(x^3-4x^2+4x)=0

⇔ x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0

⇔7x^2-2=0

⇔7x^2=2

⇔7x^2=-2⇔x=-3

⇔7x^2=2⇔x=-căn 5

a, x-1:4/3+9/16=-5^(^laf mũ)2/4^2

x-1:4/3+9/16=25/16

x-1:4/3=25/16-9/16

x-1:4/3=1

x-1=1.4/3

x-1=4/3

x=4/3+1

x=7/3

kl:...................(hì tự viết nha mik lười viết)

b, 5/2x+2-31/2=1/6

5/2x+2=1/6+31/2

5/2x+2=16

5/2x=16-2

5/2x=14

x=14:5/2

x=28/5(vì 28 ko chia hết cho 5 nên mik viết như vậy)

*chúc bn học tốt nha*         ^-^                        *_*

c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)

\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)

\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)

\(x=-\dfrac{15}{2}\)

d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)

\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)

A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)

\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)

\(x\)\(=\dfrac{11}{9}\)

B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)

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

\(x=\)\(\dfrac{-2}{3}\)

a)

\(\frac{4}{9} + x = \frac{5}{3}\)

=> \(x = \frac{5}{3}-\frac{4}{9}\)

=> \(x = \) \(\frac{11}{9}\)

Vậy \(x = \dfrac{11}{9}\)

b) 

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

=> \(x = \dfrac{-1}{2} : \dfrac{3}{4}\)

=> \(x = \dfrac{-2}{3}\)

Vậy \(x = \dfrac{-2}{3}\)

c)

\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)

=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)

=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)

=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)

=> \(x = \dfrac{-15}{2}\)

Vậy \(x = \dfrac{-15}{2}\)

d) 

\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)

=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)

=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)

Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }