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24 tháng 6

\(\frac{x}{15}=\frac{8}{21}\)

=>\(x.21=8.15\)

=>\(x.21=120\)

=>\(x\) = 120 : 21

=>\(x\) = \(\frac{40}{7}\)

Vậy \(x=\frac{40}{7}\)

24 tháng 6

\(\frac{x}{15}=\frac{8}{21}\)

\(\Rightarrow21x=15\cdot8\)

\(21x=120\)

\(x=120:21\)

\(x=\frac{120}{21}=\frac{40}{7}\)

Vậy \(x=\frac{40}{7}\)

a) Ta có: \(\dfrac{-11}{15}< \dfrac{x}{15}< \dfrac{-8}{15}\)

nên -11<x<-8

hay \(x\in\left\{-10;-9\right\}\)

b) Ta có: \(\dfrac{3}{7}< \dfrac{x}{21}< \dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{9}{21}< \dfrac{x}{21}< \dfrac{14}{21}\)

Suy ra: 9<x<14

hay \(x\in\left\{10;11;12;13\right\}\)

c) Ta có: \(\dfrac{-67}{21}< \dfrac{x}{168}< \dfrac{-3}{8}\)

nên \(\dfrac{-536}{168}< \dfrac{x}{168}< \dfrac{-63}{168}\)

Suy ra: -536<x<-63

hay \(x\in\left\{-535;-534;...;-64\right\}\)

a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)

b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)

=9/17+8/17=1

c: =>x-3/10=7/15*1/5=7/75

=>x=7/75+3/10=59/150

4 tháng 2 2018

a)    \(\frac{3}{16}+\frac{4}{15}+\frac{5}{16}+\frac{1}{15}\)

\(=\left(\frac{3}{16}+\frac{5}{16}\right)+\left(\frac{4}{15}+\frac{1}{15}\right)\)

\(=\frac{1}{2}+\frac{1}{3}\)

\(=\frac{5}{6}\)

b)   \(\frac{6}{7}\times\frac{8}{15}\times\frac{7}{6}\times\frac{15}{16}\)

\(=\left(\frac{6}{7}\times\frac{7}{6}\right)\times\left(\frac{8}{15}\times\frac{15}{16}\right)\)

\(=1\times\frac{1}{2}=\frac{1}{2}\)

c)   \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)

\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)

\(=\frac{247}{420}+\frac{72}{420}\)

\(=\frac{319}{420}\)

4 tháng 2 2018

còn phần D đâu bn 

12 tháng 8 2023

\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)

\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)

\(x-2023=1\)

\(x=2024\)

Vậy..............

12 tháng 8 2023

\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)

\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)

15 tháng 1 2021

\(x-7=-15+21\\ \Rightarrow x=-15+21+7\\ \Rightarrow x=13\)

Vậy \(x=3\)

\(8-x=-18+36\\ \Rightarrow-x=-18+36-8\\ \Rightarrow-x=10\\ \Rightarrow x=-10\)

Vậy \(x=-10\)

15 tháng 1 2021

x - 7 = -15 + 21                              8 - x = -18 + 36

x - 7 = 6                                         8 - x = 18 

x      = 7 + 6                                        x = 8 - 18

x      = 13                                             x = -10 

24 tháng 2 2022

\(\dfrac{26}{81}+\dfrac{4}{27}=\dfrac{26+12}{81}=\dfrac{38}{81}\)

\(\dfrac{5}{64}+\dfrac{7}{8}=\dfrac{5+56}{64}=\dfrac{61}{64}\)

24 tháng 2 2022

\(\dfrac{26}{81}+\dfrac{4}{27}=\dfrac{26}{81}+\dfrac{4\times3}{27\times3}=\dfrac{26}{81}+\dfrac{12}{81}=\dfrac{38}{81}\)

\(\dfrac{5}{64}+\dfrac{7}{8}=\dfrac{5}{64}+\dfrac{7\times8}{8\times8}=\dfrac{5}{64}+\dfrac{56}{64}=\dfrac{61}{64}\)

5 tháng 1 2017

A, 34+(9-21)=3417-(x+3417)

  34 + (-12) = 3417 - x - 3417

   22 = ( 3417 - 3417 ) - x

   22 = 0 - x 

     x = 0 - 22

     x = -22

 Vậy: x = - 22

B,  (15-x)+(x-21)=7-(-8+x)

 15 - x + x - 21 = 7 + 8 - x

-x + x = 7 + 8 - x - 15 + 21

0 = 15 - x - 15 + 21 

0 = ( 15 - 15 ) - x + 21

0 = 0 - x + 21

-x = 0 - 0 - 21

- x = -21

x = 21

Vậy : x = 21

5 tháng 1 2017

k mình nha, thank you!

28 tháng 10 2019

1.

b) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x.1+3^x.3^2=2430\)

\(\Rightarrow3^x.\left(1+3^2\right)=2430\)

\(\Rightarrow3^x.10=2430\)

\(\Rightarrow3^x=2430:10\)

\(\Rightarrow3^x=243\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)

Chúc bạn học tốt!