Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)

dad

a) Ta có: \(\frac{x}{9}=\frac{-12}{27}\)

=> \(27.x=-12.9\)

=> \(27x=-108\)

=> \(x=108:27\)

=>\(x=4\)

a) x=-3/20

b) x=-5/7

a)  \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)

\(\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11-8}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{5}{20}-\dfrac{8}{20}\)

\(\Leftrightarrow x=\dfrac{-3}{20}\)

Vậy x= \(\dfrac{-3}{20}\)

b)  \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{8-15}{20}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Leftrightarrow x=\dfrac{1}{4}.\dfrac{-20}{7}\)

\(\Leftrightarrow x=\dfrac{-5}{7}\)

Vậy x= \(\dfrac{-5}{7}\)

a) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow x+\dfrac{2}{5}=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{11}{12}-\dfrac{8}{12}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{5}{20}-\dfrac{8}{20}=\dfrac{-3}{20}\)

b) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{-7}{20}\)

hay \(x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{1}{4}\cdot\dfrac{-20}{7}=\dfrac{-5}{7}\)

\(a,\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{11}{12}\)

\(\Leftrightarrow x=-\dfrac{11}{12}:\dfrac{11}{12}\)

\(\Leftrightarrow x=-\dfrac{11}{12}.\dfrac{12}{11}\)

\(\Leftrightarrow x=-1\)

\(b,3-\left(\dfrac{1}{6}-x\right).\dfrac{2}{3}=\dfrac{2}{3}\)

\(\Leftrightarrow3-\dfrac{2}{3}.\left(\dfrac{1}{6}-x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow3-\dfrac{1}{9}+\dfrac{2}{3}x=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{3}-3+\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{20}{9}\)

\(\Leftrightarrow x=-\dfrac{20}{9}:\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{10}{3}\)

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{3}{20}\)

2) \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)

\(\Leftrightarrow4x=-\dfrac{20}{7}\)

\(\Leftrightarrow x=-\dfrac{5}{7}\)

a. 7/12 - 1/4 : ( -4/11 - x ) = 125%

=> 1/4 : ( -4/11 - x ) = 7/12 - 5/4 = -2/3

=> -4/11 - x = 1/4 : -2/3 

=> -4/11 - x = -3/8

=> x = -4/11 - ( -3/8 )

=> x = 1/88

b. 1/2x - 1/3 = 2/3x - 5/6

=> 1/2x - 2/3x = 1/3 - 5/6

=> -1/6x = -1/2

=> x = -1/2 : -1/6

=> x = 3

a)12(x-1)=0

=>x-1=0

=>x=1

b) 3/5x-3/4=(-3)²-|-11|

3/5x-3/4=9-11=-2

3/5x=-2+3/4=-5/4

=>x=-5/4:3/5=-25/12

c)4(x-5/8)-3/4=0,25

4(x-5/8)=0,25+3/4=1

=>x-5/8=1/4

x=1/4+5/8=7/8

2) Số đó là

27:3/5=45