\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)

\(\Leftrightarrow\dfrac{1680.\left(x-3\right)+1560.\left(x-3\right)-1456.\left(x-3\right)-1365.\left(x-3\right)}{21840}=0\)

\(\Leftrightarrow\left(x-3\right).\left(1680+1560-1456-1365\right)=0\)

\(\Leftrightarrow\left(x-3\right).419=0\)

\(\Leftrightarrow419x=1257\)

\(\Leftrightarrow x=3\)

Lời giải:

\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)

\((x-3)\left(\frac{1}{13}+\frac{1}{14}\right)=(x-3)\left(\frac{1}{15}+\frac{1}{16}\right)\)

\((x-3)\left[\left(\frac{1}{13}+\frac{1}{14}\right)-\left(\frac{1}{15}+\frac{1}{16}\right)\right]=0\)

Ta thấy:

\(\frac{1}{13}>\frac{1}{15}; \frac{1}{14}>\frac{1}{16}\Rightarrow \frac{1}{13}+\frac{1}{14}> \frac{1}{15}+\frac{1}{16}\)

Do đó biểu thức trong ngoặc vuông lớn hơn $0$ hay khác $0$

$\Rightarrow x-3=0$

$\Leftrightarrow x=3$

a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)

\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)

hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)

b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)

\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)

c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)

\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)

\(\Leftrightarrow x\in\left\{1;2;3\right\}\)

ko biết =))

Giải:

\(9-3\times\left(x-9\right)=6\) 

      \(3\times\left(x-9\right)=9-6\) 

      \(3\times\left(x-9\right)=3\) 

               \(x-9=3:3\) 

               \(x-9=1\) 

                     \(x=1+9\) 

                     \(x=10\) 

\(4+6\times\left(x+1\right)=70\) 

      \(6\times\left(x+1\right)=70-4\) 

      \(6\times\left(x+1\right)=66\) 

               \(x+1=66:6\) 

               \(x+1=11\) 

                     \(x=11-1\) 

                     \(x=10\) 

\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\) 

         \(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\) 

         \(\dfrac{x}{13}=\dfrac{4}{13}\) 

\(\Rightarrow x=4\) 

\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\) 

         \(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\) 

         \(\dfrac{3}{x}=\dfrac{3}{7}\) 

\(\Rightarrow x=7\) 

\(5\times\left(3+7\times x\right)=40\) 

         \(3+7\times x=40:5\) 

         \(3+7\times x=8\) 

                \(7\times x=8-3\) 

                \(7\times x=5\) 

                      \(x=5:7\) 

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

\(x\times6+12:3=120\) 

       \(x\times6+4=120\) 

             \(x\times6=120-4\) 

             \(x\times6=116\) 

                   \(x=116:6\) 

                   \(x=\dfrac{58}{3}\) 

\(x\times3,7+x\times6,3=120\) 

    \(x\times\left(3,7+6,3\right)=120\) 

                  \(x\times10=120\) 

                           \(x=120:10\) 

                           \(x=12\) 

\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\) 

      \(\left(360-x\right):0,25=400\) 

                   \(360-x=400.0,25\) 

                   \(360-x=100\) 

                             \(x=360-100\) 

                             \(x=260\) 

\(71+65\times4=\dfrac{x+140}{x}+260\) 

\(\left(x+140\right):x+260=71+260\) 

\(x:x+140:x+260=331\) 

    \(1+140:x+260=331\) 

                    \(140:x=331-1-260\) 

                    \(140:x=70\) 

                             \(x=140:70\) 

                             \(x=2\) 

\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\) 

                      \(10\times x+\left(1+4+7+...+28\right)=155\)

Số số hạng \(\left(1+4+7+...+28\right)\) :

         \(\left(28-1\right):3+1=10\) 

Tổng dãy \(\left(1+4+7+...+28\right)\) :

         \(\left(1+28\right).10:2=145\) 

\(\Rightarrow10\times x+145=155\) 

               \(10\times x=155-145\) 

               \(10\times x=10\) 

                       \(x=10:10\) 

                       \(x=1\) 

Đều theo cách lớp 5 nha em!

`#3107`

`7/13 \div x = 14/39`

`=> x = 7/13 \div 14/39`

`=> x = 3/2`

Vậy, `x = 3/2`

________

`x \times 3/5 = 14/15`

`=> x = 14/15 \div 3/5`

`=> x = 14/9`

Vậy, `x=14/9.`

\(\dfrac{7}{13}:x=\dfrac{14}{39}\)

\(x=\dfrac{7}{13}:\dfrac{14}{39}\)

\(x=\dfrac{7}{13}\times\dfrac{39}{14}\)

\(x=\dfrac{3}{2}\)

___

\(x\times\dfrac{3}{5}=\dfrac{14}{15}\)

\(x=\dfrac{14}{15}:\dfrac{3}{5}\)

\(x=\dfrac{14}{15}\times\dfrac{5}{3}\)

\(x=\dfrac{14}{9}\)

` 7/13 : x = 14/39 `

`           x = 7/13 : 14/39 `

`           x =  7/13 xx 39/14`

`          x = 3/2`

Vậy ` x=  3/2.`

` x xx 3/5 = 14/15 `

` x = 14/15 : 3/5 `

` x = 14/15 xx 5/3 `

` x = 14/9` 

Vậy ` x = 14/9.`

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

\(\dfrac{15-x}{14}=\dfrac{5}{7}\\ \left(15-x\right)\times7=5\times14\\ \left(15-x\right)\times7=70\\ 15-x=70:7\\ 15-x=10\\ x=15-10\\ x=5\)

__

\(\dfrac{x}{9}=\dfrac{17}{3}\\ x\times3=17\times9\\ x\times3=153\\ x=153:3\\ x=51\)

__

\(15-\dfrac{x}{2}=\dfrac{3}{7}\\ \dfrac{x}{2}=15-\dfrac{3}{7}\\ \dfrac{x}{2}=\dfrac{102}{7}\\ x\times7=102\times2\\ x\times7=204\\ x=204:7\\ x=\dfrac{204}{7}\)

???

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

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

a: =>x-5/14=6/14-1/14=5/14

=>x=10/14=5/7

b; =>2/9:x=3/6+4/6=7/6

=>x=2/9:7/6=2/9*6/7=4/21

c: =>32:x=8

=>x=4

a

\(x-\dfrac{5}{14}=\dfrac{3}{7}-\dfrac{1}{14}\\ x-\dfrac{5}{14}-\dfrac{3}{7}+\dfrac{1}{14}=0\\ x+\dfrac{1}{14}-\dfrac{5}{14}-\dfrac{6}{14}=0\\ x+\dfrac{1-5-6}{14}=0\\ x-\dfrac{5}{7}=0\\ x=0+\dfrac{5}{7}\\ x=\dfrac{5}{7}\)

b

\(\dfrac{2}{9}:x=\dfrac{1}{2}+\dfrac{2}{3}\\ \dfrac{2}{9}:x=\dfrac{3}{6}+\dfrac{4}{6}\\ \dfrac{2}{9}:x=\dfrac{3+4}{6}=\dfrac{7}{6}\\ x=\dfrac{2}{9}:\dfrac{7}{6}\\ x=\dfrac{2}{9}\times\dfrac{6}{7}=\dfrac{2.3.2}{3.3.7}=\dfrac{4}{21}\)

c

\(\dfrac{6}{32:x}=\dfrac{12}{16}\\ 32:x=6:\dfrac{12}{16}\\ 32:x=6\times\dfrac{16}{12}\\ 32:x=\dfrac{3\times2\times4\times4}{3\times4}\\ 32:x=8\\ x=\dfrac{32}{8}\\ x=4\)

`@` `\text {Answer}`

`\downarrow`

`a,`

`x - 5/14 = 3/7 - 1/14`

`x - 5/14 = 5/14`

`=> x = 5/14 + 5/14`

`=> x = 5/7`

Vậy, `x = 5/7`

`b,`

`2/9 \div x = 1/2 + 2/3`

`2/9 \div x = 7/6`

`x = 2/9 \div 7/6`

`x = 4/21`

Vậy, `x = 4/21`

`c,`

\(\dfrac{6}{32\div x}=\dfrac{12}{16}\)

`6/(32 \div x) = 3/4`

`32 \div x = 6 \div 3/4`

`32 \div x = 8`

` x = 32 \div 8`

`x = 4`

Vậy, `x = 4`