Bài 3:

a) 

\(\dfrac{2}{3}x+4=-12\\ \Rightarrow\dfrac{2}{3}x=-12-4=-16\\ \Rightarrow x=-16:\dfrac{2}{3}\Rightarrow x=-24\)

b) \(\dfrac{5}{6}-\dfrac{1}{6}:x=-2\dfrac{1}{2}\)

\(\Rightarrow\dfrac{5}{6}-\dfrac{1}{6}:x=-\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{6}:x=\dfrac{5}{6}+\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{6}:x=\dfrac{14}{6}\\ \Rightarrow x=\dfrac{1}{6}:\dfrac{14}{6}=\dfrac{1}{14}\) 

c) \(x:20-25\%x=-1\dfrac{1}{5}\)

\(\Rightarrow\dfrac{x}{20}-\dfrac{x}{4}=-\dfrac{4}{5}\)

\(\Rightarrow\dfrac{x}{20}-\dfrac{5x}{20}=-\dfrac{4}{5}\)

\(\Rightarrow-\dfrac{4x}{20}=\dfrac{-4}{5}\)

\(\Rightarrow-\dfrac{x}{5}=-\dfrac{4}{5}\)

\(\Rightarrow x=-4\)

d) \(\dfrac{35}{x-1}=\dfrac{15}{-6}\left(x\ne1\right)\)

\(\Rightarrow-6\cdot35=15\left(x-1\right)\\ \Rightarrow-210=15x-15\\ \Rightarrow15x=-210+15=-195\\ \Rightarrow x=\dfrac{-195}{15}\\ \Rightarrow x=-13\)

e) 

\(\left(\dfrac{9}{25}\right)^x=\left(\dfrac{5}{3}\right)^{-1}\cdot\left(\dfrac{3}{5}\right)^5\\ \Rightarrow\left[\left(\dfrac{3}{5}\right)^2\right]^x=\dfrac{3}{5}\cdot\left(\dfrac{3}{5}\right)^5\\ \Rightarrow\left(\dfrac{3}{5}\right)^{2x}=\left(\dfrac{3}{5}\right)^6\\ \Rightarrow2x=6\\ \Rightarrow x=3\)

f) 

\(0,5^{x+1}+0,5^x=1,5\\ \Rightarrow0,5^x\cdot\left(0,5+1\right)=1,5\\ \Rightarrow0,5^x\cdot1,5=1,5\\ \Rightarrow0,5^x=1,5:1,5=1\\ \Rightarrow0,5^x=0,5^0\\ \Rightarrow x=0\)

Bài 3:

a: \(\dfrac{2}{3}x+4=-12\)

=>\(\dfrac{2}{3}x=-12-4=-16\)

=>\(x=-16:\dfrac{2}{3}=-16\cdot\dfrac{3}{2}=-24\)

b: \(\dfrac{5}{6}-\dfrac{1}{6}:x=-2\dfrac{1}{2}\)

=>\(\dfrac{5}{6}-\dfrac{1}{6}:x=-\dfrac{5}{2}\)

=>\(\dfrac{1}{6}:x=\dfrac{5}{6}+\dfrac{5}{2}=\dfrac{5}{6}+\dfrac{15}{6}=\dfrac{20}{6}=\dfrac{10}{3}\)

=>\(x=\dfrac{1}{6}:\dfrac{10}{3}=\dfrac{1}{6}\cdot\dfrac{3}{10}=\dfrac{1}{20}\)

c: \(x:20-25\%\cdot x=-1\dfrac{1}{5}\)

=>\(0,05x-0,25x=-1,2\)

=>-0,2x=-1,2

=>x=1,2:0,2=6

d: \(\dfrac{35}{x-1}=\dfrac{15}{-6}\)(ĐKXĐ: \(x\ne1\))

=>\(x-1=\dfrac{35\cdot\left(-6\right)}{15}=\dfrac{-210}{15}=-14\)

=>x=-14+1=-13(nhận)

e: \(\left(\dfrac{9}{25}\right)^x=\left(\dfrac{5}{3}\right)^{-1}\cdot\left(\dfrac{3}{5}\right)^5\)

=>\(\left(\dfrac{3}{5}\right)^{2x}=\left(\dfrac{3}{5}\right)\cdot\left(\dfrac{3}{5}\right)^5=\left(\dfrac{3}{5}\right)^6\)

=>2x=6

=>x=3

f: \(0,5^{x+1}+0,5^x=1,5\)

=>\(0,5^x\cdot\left(0,5+1\right)=1,5\)

=>\(0,5^x=1\)

=>x=0

Bài 2:

a: \(\dfrac{-23}{32}+\dfrac{14}{21}+\dfrac{-9}{32}+\dfrac{28}{21}\)

\(=\left(-\dfrac{23}{32}-\dfrac{9}{32}\right)+\left(\dfrac{14}{21}+\dfrac{28}{21}\right)\)

\(=-\dfrac{32}{32}+\dfrac{42}{21}=-1+2=1\)

b: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)\)

\(=\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{3}{12}\right)\cdot\left(\dfrac{16}{20}-\dfrac{15}{20}\right)\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{20}=\dfrac{17}{240}\)

c: \(\dfrac{-5}{7}\cdot16\dfrac{1}{3}+\dfrac{5}{7}\cdot\left(-23\dfrac{2}{3}\right)\)

\(=\dfrac{-5}{7}\cdot\left(16+\dfrac{1}{3}+23+\dfrac{2}{3}\right)\)

\(=-\dfrac{5}{7}\cdot40=-\dfrac{200}{7}\)

d: \(6\dfrac{4}{9}:\dfrac{7}{2}+7\dfrac{5}{9}:\left(\dfrac{2}{7}\right)^{-1}-\dfrac{1}{2}\)

\(=\left(6+\dfrac{4}{9}\right)\cdot\dfrac{2}{7}+\left(7+\dfrac{5}{9}\right)\cdot\dfrac{2}{7}-\dfrac{1}{2}\)

\(=\dfrac{2}{7}\left(6+\dfrac{4}{9}+7+\dfrac{5}{9}\right)-\dfrac{1}{2}=\dfrac{2}{7}\cdot14-\dfrac{1}{2}=4-\dfrac{1}{2}=\dfrac{7}{2}\)

e: \(\left(2^3:\dfrac{1}{2}\right)\cdot\dfrac{1}{8}+\dfrac{1}{9}\cdot\left(-3\right)^2-\left(-\dfrac{1}{2015}\right)^0\)

\(=\left(8\cdot2\right)\cdot\dfrac{1}{8}+\dfrac{1}{9}\cdot9-1=2+1-1=2\)