\(\dfrac{x+1}{15}+\dfrac{x+2}{14}=\dfrac{x+3}{13}+\dfrac{x+4}{12}\\ \Rightarrow\left(\dfrac{x+1}{15}+1\right)+\left(\dfrac{x+2}{14}+1\right)-\left(\dfrac{x+3}{13}+1\right)-\left(\dfrac{x+4}{12}+1\right)=0\\ \dfrac{x+16}{15}+\dfrac{x+16}{14}-\dfrac{x+16}{13}-\dfrac{x+16}{12}=0\\ \left(x+16\right)\left(\dfrac{1}{15}+\dfrac{1}{14}-\dfrac{1}{13}-\dfrac{1}{12}\right)=0\\ x+16=0\\ x=-16\)

a, \(x\) : \(\dfrac{13}{3}\) = -2,5

    \(x\)         = -2,5 . \(\dfrac{13}{3}\)

    \(x\)         = \(\dfrac{65}{6}\)

b,\(\dfrac{3}{5}\)\(x\)         = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)

   \(\dfrac{3}{5}x\)         = \(\dfrac{-3}{20}\)

      \(x\)          = \(\dfrac{-3}{20}\) :  \(\dfrac{3}{5}\)

      \(x\)          = \(\dfrac{-1}{4}\)

c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)

              \(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)

               \(\dfrac{12}{13}x=2\)

                    \(x=2:\dfrac{12}{13}\)

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

a) \(\dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-5}{9}+\dfrac{5}{12}=\dfrac{-20}{36}+\dfrac{15}{36}=-\dfrac{5}{36}\)

b) \(\dfrac{-5}{12}:\dfrac{15}{4}=\dfrac{-5}{12}\times\dfrac{4}{15}=\dfrac{-1}{9}\)

c) \(\dfrac{1}{13}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{13}-\dfrac{14}{13}=\dfrac{1}{13}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{14}{13}=\dfrac{1}{13}\cdot1-\dfrac{14}{13}=\dfrac{1}{13}-\dfrac{14}{13}=-1\)

a)=--5/36

b)=-1/9

c) =-1 

cách làm như trên

\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)

\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)

Vậy \(x\in\left\{22;23;24;...\right\}\)

\(\dfrac{????????}{????????????}\)

a) \(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

   \(\dfrac{11}{55}+\dfrac{10}{55}< \dfrac{x}{55}< \dfrac{22}{55}+\dfrac{1}{55}\)

   \(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{23}{55}\)

\(\Rightarrow\) \(x=22\)

b) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)

  \(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}< x\le\dfrac{26}{8}+\dfrac{14}{8}\)

  \(1< x\le5\)

  \(\Rightarrow\) \(x\in\) {\(2;3;4;5\)}

c) \(\dfrac{1}{3}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\)

 Ko biết làm

d) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)

   \(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)

   \(4\le x\le9\)

   \(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}

x = 13/12 - 3/4

x = 1/3

x=13/12-3/4

x=13/12-9/12

x=4/12

\(x+\dfrac{3}{4}=\dfrac{13}{12}\)

\(\Rightarrow x=\dfrac{13}{12}-\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{13}{12}-\dfrac{9}{12}\)

\(\Rightarrow x=\dfrac{1}{3}\)

a: \(=\dfrac{7+12-6}{13}=1\)

b: \(=\dfrac{13}{10}\cdot\dfrac{6-26}{13}=\dfrac{-20}{10}=-2\)

c: \(=\dfrac{3}{4}\cdot2-\dfrac{5}{2}\cdot\dfrac{-4}{3}=\dfrac{3}{2}+\dfrac{20}{6}=\dfrac{3}{2}+\dfrac{10}{3}=\dfrac{29}{6}\)

d: \(=\dfrac{3}{8}\cdot\dfrac{8}{5}+\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)

Bài 4. Trên tia Ox lấy điểm A và B sao cho OA = 8cm, OB = 12cm.

a, Điểm A có nằm giữa hai điểm O và B không? Vì sao? Tính AB.

b, Gọi M, N lần lượt là trung điểm của OA, OB. Điểm M có nằm giữa hai điểm O và N không? Vì sao? Tính MN.

c, Điểm N có là trung điểm của đoạn thẳng AM không? Vì sao? 
cho mìn hỏi câu b nhoa

3/7x2=6/7

5/12x1=5/12

3x4/7=12/7

0x5/9=0

có dc tính nhanh ko:)?

a) =  2/4 + 1/24 = 13/24

b) = 1/2 x ( 3/5 + 12/5 ) = 1/2 x 3 = 3/2

2/4 + 1/24 = 13/24

3/10 + 6/5 = 3/2

a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)

\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)

\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)

\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)

mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)

nên x-17=0

hay x=17

Vậy: x=17

b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)

\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)

\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)

nên x+20=0

hay x=-20

Vậy: x=-20