3\(x\) - \(\dfrac{12}{5}\) = -0,6

3\(x\) - 2,4  = -0,6

3\(x\)          = 0,6 + 2,4

3\(x\)          = 3

   \(x\)         = 1

Cô làm rồi mà em 

TA CÓ 0=0²

⇒X-11+Y+X+4-Y=0

⇒(X+X)+(-11+4)+(Y-Y)=0

⇒2X+(-7)+0=0

⇒2X=0-(-7)

⇒2X=7

⇒X=7:2

⇒X=3,5

VẬY X =3,5

112.

a) (a + b)(c + d) = ac + ad + bc + bd

b) (a - b)(c - d) = ac - ad - bc + bd

113.

a) \(A=\dfrac{\dfrac{-6}{5}+\dfrac{6}{19}-\dfrac{6}{23}}{\dfrac{9}{5}-\dfrac{9}{19}+\dfrac{9}{23}}=\dfrac{-6\left(\dfrac{1}{5}-\dfrac{1}{19}+\dfrac{1}{23}\right)}{9\left(\dfrac{1}{5}-\dfrac{1}{19}+\dfrac{1}{23}\right)}=\dfrac{-2}{3}\)

b) \(B=\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}{\dfrac{1}{4}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}=\dfrac{4}{3}\)

114. Để \(A=\dfrac{2n+7}{n+1}\) là một số nguyên thì \(\left(2n+7\right)⋮\left(n+1\right)\). Suy ra

\(\left(2n+7\right)-2\left(n+1\right)⋮\left(n+1\right)\)

\(5⋮\left(n+1\right)\)

\(\left(n+1\right)\inƯ\left(5\right)\)

\(\left(n+1\right)\in\left\{1;5;-1;-5\right\}\)

\(n\in\left\{0;4;-2;-6\right\}\)

Vậy để A là một số nguyên thì \(n\in\left\{0;4;-2;-6\right\}\)

mn nhanh lên ạaa!!!

Đề bài cụ thể là gì vậy ạ

trả lời giúp mik

đội 1  :48:(7-3)x7= 84

đội 2  :84-48=36                                     

a,     \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\)    = \(\dfrac{10}{21}\)

    (\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\)  =  \(\dfrac{10}{21}\)

     - \(\dfrac{5}{21}\) \(\times\) \(x\)      = \(\dfrac{10}{21}\)

                 \(x\)      = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))

                 \(x\)      = -2 

b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)

            \(x\) - \(\dfrac{1}{3}\)    =  \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))

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

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

              \(x\)      = - \(\dfrac{13}{6}\)

c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)

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

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

              \(x\)           = - \(\dfrac{47}{10}\)

\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)

\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)

Bài 17:

\(\widehat{G_3}\) = \(\widehat{G_1}\) = 115⁰ (hai góc đối đỉnh)

\(\widehat{G_2}\) = \(\widehat{G_4}\) = 180⁰ - 115⁰ = 65⁰

\(\widehat{H_4}\) = \(\widehat{H_1}\) = 115⁰ (hai góc đối đỉnh)

\(\widehat{H_2}\) = \(\widehat{H_3}\) = \(\widehat{G_4}\) = 65⁰

a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)

\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)

b) 

\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)

\(=-14\)

c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)

\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)

Chu vi đáy là:

  \(6.3=18\) ( dm )

Chiều cao cột mốc là:

   \(270:18=15\) ( dm )

                Đ/S:...

Chu vi đáy:

\(6\cdot3=18\left(dm^2\right)\)

Chiều cao cột mốc:

\(270:18=15\left(dm\right)\)

Đáp số: 15dm