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30 tháng 6 2017

\(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)

\(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)

\(\Leftrightarrow\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)

\(\Leftrightarrow\left(123-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)

\(\Leftrightarrow123-x=0\Leftrightarrow x=123\)

Vậy x = 123

30 tháng 6 2017

thanks you

27 tháng 10 2017

ai bít

a: \(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}-10=0\)

\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)

=>2007-x=0

hay x=2007

b: \(\Leftrightarrow x+\left(1+1+1+1+1+1+1\right)+\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=0\)

\(\Leftrightarrow x+7+\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=0\)

=>x+7+1/3-1/10=0

hay x=-217/30

5 tháng 7 2017

\(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)

\(\Rightarrow\dfrac{x-241}{17}-1+\dfrac{x-220}{19}-2+\dfrac{x-195}{21}-3+\dfrac{x-166}{23}-4=0\)

\(\Rightarrow\dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-258}{21}+\dfrac{x-258}{23}=0\)

\(\Rightarrow\left(x-258\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0\)

\(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\ne0\)

\(\Rightarrow x-258=0\Rightarrow x=258\)

Vậy x = 258

14 tháng 11 2018

x−24117+x−22019+x−19521+x−16623=10x−24117+x−22019+x−19521+x−16623=10

⇒x−24117−1+x−22019−2+x−19521−3+x−16623−4=0⇒x−24117−1+x−22019−2+x−19521−3+x−16623−4=0

⇒x−25817+x−25819+x−25821+x−25823=0⇒x−25817+x−25819+x−25821+x−25823=0

⇒(x−258)(117+119+121+123)=0⇒(x−258)(117+119+121+123)=0

117+119+121+123≠0117+119+121+123≠0

⇒x−258=0⇒x=258

30 tháng 6 2023

a) Ta có:

\(A=\dfrac{-68}{123}\cdot\dfrac{-23}{79}=\dfrac{68}{123}\cdot\dfrac{23}{79}\)

\(B=\dfrac{-14}{79}\cdot\dfrac{-68}{7}\cdot\dfrac{-46}{123}=-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)\)

\(C=\dfrac{-4}{19}\cdot\dfrac{-3}{19}\cdot...\cdot\dfrac{0}{19}\cdot...\cdot\dfrac{3}{19}\cdot\dfrac{4}{19}=0\)

Suy ra A là số hữu tỉ dương, B là số hữu tỉ âm và C là 0.

Vậy A > C > B.

b) Ta có:

\(\dfrac{B}{A}=\dfrac{-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)}{\dfrac{68}{123}\cdot\dfrac{23}{79}}=-\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\cdot\dfrac{123}{68}\cdot\dfrac{79}{23}\)

\(\dfrac{B}{A}=-\dfrac{14\cdot68\cdot46\cdot123\cdot79}{79\cdot7\cdot123\cdot68\cdot23}=-\left(2\cdot2\right)=-4\)

Vậy B : A = -4

18 tháng 9 2023

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 - \(\dfrac{1}{5}\)

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

18 tháng 9 2023

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

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

a) Ta có: \(\left|2.5-x\right|=1.3\)

\(\Leftrightarrow\left|x-2.5\right|=1.3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2.5=1.3\\x-2.5=-1.3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.8\\x=-1.3+2.5=1.2\end{matrix}\right.\)

Vậy: \(x\in\left\{3.8;1.2\right\}\)

b) Ta có: \(1.6-\left|x-0.2\right|=0\)

\(\Leftrightarrow\left|x-0.2\right|=1.6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0.2=1.6\\x-0.2=-1.6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.4\end{matrix}\right.\)

Vậy: \(x\in\left\{1.8;-1.4\right\}\)

c) Ta có: \(13^x=169\)

\(\Leftrightarrow13^x=13^2\)

\(\Leftrightarrow x=2\)

Vậy: x=2

d) Ta có: \(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\)

\(\Leftrightarrow\dfrac{2}{x}=\dfrac{x}{\dfrac{8}{25}}\)

\(\Leftrightarrow x^2=\dfrac{16}{25}\)

hay \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

Vậy: \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

7 tháng 2 2021

\(|2,5-x|=1,3\)

\(\Rightarrow2,5-x=1,3\)    hoặc      -(2,5-x)=1,3

=>x=2,5-1,3                          x-2,5=1,3

=>x=1,2                                x=1,3+2,5=3,8

Vậy \(x\in\left\{1,2;3,8\right\}\)

\(1,6-|x-0,2|=0\Rightarrow|x-0,2|=1,6\)

=> x-0,2=1,6       hoặc       -(x-0,2)=1,6

=>x=1,6+0,2                      0,2-x=1,6

=>x=3,8                              x=0,2-1,6=-1,4

Vậy \(x\in\left\{3,8;-1,4\right\}\)

\(13^x=169\Rightarrow13^x=13^2\Rightarrow x=2\)

\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\Rightarrow-2\times\dfrac{8}{25}=-x\times x\Rightarrow\dfrac{-16}{25}=-x^2\Rightarrow\dfrac{16}{25}=x^2\Rightarrow x^2=\left(\dfrac{4}{5}\right)^2=\left(-\dfrac{4}{5}\right)^2\)

\(\Rightarrow x=\dfrac{4}{5}\) hoặc \(x=-\dfrac{4}{5}\)