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31 tháng 8 2016

\(\frac{x-11}{95}+\frac{x-13}{93}=\frac{x-15}{91}+\frac{x-17}{89}\) => \(\frac{x-11}{95}-1+\frac{x-13}{93}-1=\frac{x-15}{91}-1+\frac{x-17}{89}-1\)

=>\(\frac{x-106}{95}+\frac{x-106}{93}=\frac{x-106}{91}+\frac{x-106}{89}\)

=>\(\left(\frac{1}{95}+\frac{1}{93}\right)\left(x-106\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\left(x-106\right)=0\)

<=>\(\left[\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\right]\left(x-106\right)=0\).Vì\(\frac{1}{95}< \frac{1}{91};\frac{1}{93}< \frac{1}{89}\) nên\(\frac{1}{95}+\frac{1}{93}< \frac{1}{91}+\frac{1}{89}\)

=>\(\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)< 0\) hay khác 0.Vậy x - 106 = 0, tìm được x = 106

8 tháng 10 2023

\(5x-\dfrac{4}{5}=x+\dfrac{13}{15}\)

\(\Rightarrow5x-x=\dfrac{13}{15}+\dfrac{4}{5}\)

\(\Rightarrow4x=\dfrac{13}{15}+\dfrac{12}{15}\)

\(\Rightarrow4x=\dfrac{25}{15}\)

\(\Rightarrow4x=\dfrac{5}{3}\)

\(\Rightarrow x=\dfrac{5}{3}:4\)

\(\Rightarrow x=\dfrac{5}{12}\)

8 tháng 10 2023

      5x-4/5=x+13/15

<=>5x-x=13/15+4/5

<=>4x=5/3

<=>x=5/12

vậy ...

1 tháng 8 2017

\(\dfrac{x+2}{17}+\dfrac{x+4}{15}+\dfrac{x+6}{13}=\dfrac{x+8}{11}+\dfrac{x+10}{9}+\dfrac{x+12}{7}\)

\(\Leftrightarrow\dfrac{x+2}{17}+1+\dfrac{x+4}{15}+1+\dfrac{x+6}{13}=\dfrac{x+8}{11}+1+\dfrac{x+10}{9}+1+\dfrac{x+12}{7}+1\)

\(\Leftrightarrow\dfrac{x+19}{17}+\dfrac{x+19}{15}+\dfrac{x+19}{13}=\dfrac{x+19}{11}+\dfrac{x+19}{9}+\dfrac{x+19}{7}\)

\(\Leftrightarrow\left(x+19\right)\left(\dfrac{1}{17}+\dfrac{1}{15}+\dfrac{1}{13}-\dfrac{1}{11}-\dfrac{1}{9}-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow x+19=0\)

\(\Leftrightarrow x=-19\)

Vậy ...

1 tháng 8 2017

(x+2)/17+(x+4)/15+(x+6)/13-(x+8)/11-(x+10)/9 + (x+12)/7=0

Cộng 1 vào các hạng tử ta được:

(x+19)/17+(x+19)/15+(x+19)/13-(x+19)/11-(x+19)/9-(x+19)/7=0

=> (x+19)(1/17+1/15+1/13-1/11-1/9-1/7)=0

=>x+19=0

=>x=-19

Vậy x= -19

Chúc bạn học tốt!!!

19 tháng 3 2022

\(\dfrac{x+11}{89}+\dfrac{x+13}{87}=\dfrac{x+15}{85}+\dfrac{x+17}{83}\\\Leftrightarrow\left(\dfrac{x+11}{89}+1\right)+\left(\dfrac{x+13}{87}+1\right)=\left(\dfrac{x+15}{85}+1\right)+\left(\dfrac{x+17}{83}+1\right)\\ \Leftrightarrow\dfrac{x+100}{89}+\dfrac{x+100}{87}-\dfrac{x+100}{85}-\dfrac{x+100}{83}=0\\ \Leftrightarrow\left(x+100\right)\left(\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\right)=0\\ \Leftrightarrow x+100=0\left(vì\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\ne0\right)\\ \Leftrightarrow x=-100\)