\(\dfrac{7}{3}-\dfrac{5}{6}=\dfrac{14}{6}-\dfrac{5}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)

\(0,125+y.2=18,5\)

    ⇔      \(y.2=18,375\)

    ⇔         \(y=9,1875\)

\(\dfrac{2}{5}+x:\dfrac{7}{3}=\dfrac{3}{2}\)

⇔    \(x:\dfrac{7}{3}=\dfrac{11}{10}\)

   ⇔      \(x=\dfrac{77}{30}\)

\(\dfrac{9}{4}.\dfrac{7}{3}=\dfrac{21}{4}\)

Đâu là chất gây nghiện bị Nhà Nước cấm?

A.Rượu                     

B.Bia                         

C.Thuốc lá               

 D.Ma túy

\(\dfrac{36}{21}-\dfrac{15}{20}=\dfrac{12}{7}-\dfrac{3}{4}=\dfrac{27}{28}\)

\(\dfrac{63}{45}-\dfrac{20}{25}=\dfrac{7}{5}-\dfrac{4}{5}=\dfrac{3}{5}\)

Ta có : 

\(\Delta=b^2-4.a.c\)

\(\Delta=[-\left(5-m\right)]^2-4.1.\left(4m+4\right)\)

\(\Delta=25-10m+m^2-4.\left(4m+4\right)\)

\(\Delta=25-10m+m^2-16m-16\)

\(\Delta=m^2-26m+9\)

\(\Delta=\left(m-13\right)^2-160\) > 0 \(\forall m\) \(\in R\)

Theo ht vi - ét , ta có :
\(x_1+x_2=\) \(5+m\)

\(x_1.x_2=4m+4\)

    \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{7}{12}\)

 ⇔ \(x_1+x_2=\dfrac{7}{12}\)

⇔ \(5+m=\dfrac{7}{12}\)

⇔ \(m=-\dfrac{53}{12}\)

Vậy m = \(-\dfrac{53}{12}\) 

( không chắc đáp án đâu nhé )

    \(2x-\dfrac{3}{5}=\dfrac{2}{3}\)

⇔ \(2x=\dfrac{2}{3}+\dfrac{3}{5}\)

⇔ \(2x=\dfrac{19}{15}\)

⇔ \(x=\dfrac{19}{15}:2\)

⇔ \(x=\dfrac{19}{30}\)

Vậy \(x=\dfrac{19}{30}\)

\(\dfrac{117}{234}=\dfrac{1}{2}\)

\(\dfrac{90}{126}=\dfrac{5}{7}\)

\(\dfrac{162162}{345345}=\dfrac{54}{115}\)