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11 tháng 3 2017

\(\dfrac{x-90}{10}+\dfrac{x-76}{12}=\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)

\(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)=\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)\)\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}=\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-\dfrac{1}{17}\right)=0\)

\(\Leftrightarrow x-100=0\)

\(\Rightarrow x=100\)

vậy \(S=\left\{100\right\}\)

11 tháng 3 2017

thanks

22 tháng 2 2019

\(\dfrac{x-90}{10}+\dfrac{x-76}{12}=\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)

\(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)=\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)\)\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}=\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-\dfrac{1}{17}\right)=0\)

\(\Leftrightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy \(S=\left\{100\right\}\)

b: \(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3\left(x^2+x-6\right)\)

\(\Leftrightarrow3x^2-10x+3=3x^2+3x-18\)

=>-13x=-21

hay x=21/13

c: \(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\)

=>x-100=0

hay x=100

30 tháng 1 2018

1a)\(\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)

=> \(\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)

=>\(\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)

=> x=100( vi \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\ne0\)

b) \(\dfrac{x-5}{2012}-1+\dfrac{x-4}{2013}-1=\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1\)

=> \(\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}-\dfrac{x-2017}{2015}=0\)

=>(x-2017).\(\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)

=> x=2015(vi .....................................................≠0)

2)

11 tháng 3 2017

\(\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+\dfrac{x+16}{4}=4\)

\(\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}=4-\dfrac{x+16}{4}\)

\(\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}\right)=-x\)

Mk giải đế đây rùi bạn tự giải nốt đi

11 tháng 3 2017

À bạn có chs f ko kết pạn

a: =>15/5*34/17<x<120/15

=>6<x<8

=>x=7

b; =>20/5*58/29<x<112/11

=>8<x<112/11

=>x=9 hoặc x=10

18 tháng 8 2021

\(\frac{X-90}{10}+\frac{X-76}{12}+\frac{X-58}{14}+\frac{X-36}{16}+\frac{X-15}{17}=15\)

\(\frac{X-90}{10}+\frac{X-76}{12}+\frac{X-58}{14}+\frac{X-36}{16}+\frac{X-15}{17}-15=0\)

\(\frac{X-90}{10}-1+\frac{X-76}{12}-2+\frac{X-58}{14}-3+\frac{X-36}{16}-4+\frac{X-15}{17}-5=0\)

\(\frac{X-100}{10}+\frac{X-100}{12}+\frac{X-100}{14}+\frac{X-100}{16}+\frac{X-100}{17}=0\)

\(\left(X-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

=> X-100=0

=> X=100

vậy x=100

15 tháng 3 2020

\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)

\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)

\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

có : \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)

\(\Leftrightarrow x-100=0\)

\(\Leftrightarrow x=100\)

15 tháng 3 2020

\(pt\)\(\Leftrightarrow\)\(({x-90\over10}-1)+({x-76\over12}-2)+\)\(+({x-58\over14}-3)+({x-36\over16}-4)+({x-15\over17}-5)=0\)

\(\Leftrightarrow\)\(({x-100\over10})+({x-100\over12})+({x-100\over14})+({x-100\over16})\)

\(+({x-100\over17})=0\)

\(\Leftrightarrow\)\((x-100)({1\over10}+{1\over12}+{1\over14}+{1\over16}+{1\over17})=0\)

\(\Rightarrow\)\(x-100=0\)

\(\Rightarrow\)\(x=100\)