Question

Tìm x

a) \(x-\frac{1}{3}=\frac{5}{14}.\left(\frac{-7}{6}\right)\)

b) \(\frac{3}{4}+\frac{1}{4}.x=0,2\)

c) \(\frac{1}{12}.x^2=1.\frac{1}{3}\)

Answer 1

a) x-1/3=5/14.(-7/6)

  x-1/3=-5/12

  x=-5/12+1/3

  x= -1/12

b) 3/4+1/4x=0,2

   1/4x=0,2-3/4

   1/4x=-11/20

    x=-11/20:1-4

    x=-11/5

c) 1/12.x^2=1.1/2

   1/12.x^2=1/2 

   x^2= 1/2:1/12

   x^2=6

=> x=căn bậc của 6

Chúc bn hok tốt

Answer 2

a)\(x-\frac{1}{3}=\frac{5}{14}\cdot\left(\frac{-7}{6}\right)\)

\(x-\frac{1}{3}=\frac{-5}{12}\)

\(x=\frac{-5}{12}+\frac{1}{3}\)

\(x=\frac{-5}{12}+\frac{4}{12}\)

\(x=\frac{-1}{12}\)

b)\(\frac{3}{4}+\frac{1}{4}\cdot x=0,2\)

\(\frac{3}{4}+\frac{1}{4}\cdot x=\frac{1}{5}\)

\(\frac{1}{4}\cdot x=\frac{1}{5}-\frac{3}{4}\)

\(\frac{1}{4}\cdot x=\frac{4}{20}-\frac{15}{20}\)

\(\frac{1}{4}\cdot x=\frac{-11}{20}\)

\(x=\frac{-11}{20}\cdot4\)

\(x=-\frac{11}{5}\)

c)\(\frac{1}{12}\cdot x^2=1\cdot\frac{1}{3}\)

\(\frac{1}{12}\cdot x^2=\frac{1}{3}\)

\(x^2=\frac{1}{3}\cdot12\)

\(x^2=4=\left(\pm2\right)^2\)

\(x=\pm2\)