tìm x biết x-3/2014 +x-7/2015=x-1/1008+x/2017-1
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Giải pt: \(\frac{x-3}{2014}+\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)
Giải phương trình \(\frac{x-3}{2014}+\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)
Tìm x,biết:
x+2015/5 + x+2014/6 = x+2017/3 + x+2018/2
Hướng dẫn: x+2015/5+1 + x+2014/6+1 = x+2017/3+1 + x+2018/2+1
=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2
=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)
Với x+2020=0=>x=-2020
Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí
Vậy x=-2020
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giải phương trình
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
\(\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-1}{1008}-2\right)+\left(\dfrac{x}{2017}-1\right)\)
\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{1008}-\dfrac{x-2017}{2017}=0\)
\(\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0\)
\(x-2017=0\) vì\(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\ne0\)
\(\Rightarrow x=2017\)
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\(\frac{x-3}{2014}+\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)
Hãy giải pt này
\(PT\Leftrightarrow\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-1}{1008}-2\right)+\left(\frac{x}{2017}-1\right)\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}=\frac{x-2017}{1008}+\frac{x-2017}{2017}\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}-\frac{x-2017}{1008}-\frac{x-2017}{2017}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}\right)=0\)
\(\Rightarrow x=2017\)
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\(\frac{x-3}{2014}\)-\(\frac{x-2}{2015}\)=\(\frac{x-1}{1008}\)+\(\frac{x}{2017}\)-1
gia sai đề rồi kía
tui làm ròi nên biết
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\(\frac{x-3}{2014}-\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)
=> \(\frac{x-3}{2014}-\frac{x-2}{2015}-2=\frac{x-1}{1008}+\frac{x}{2017}-1-2\)
=> \(\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-1}{1008}-2\right)+\left(\frac{x}{2017}-1\right)\)
=> \(\frac{x-2017}{2014}-\frac{x-2017}{2015}=\frac{x-2017}{1008}+\frac{x-2017}{2017}\)
=> \(\frac{x-2017}{2014}-\frac{x-2017}{2015}-\frac{x-2017}{1008}-\frac{x-2017}{2017}\)
=> \(\left(x-2017\right)\left(\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}\right)=0\)
Vì \(\frac{1}{1008}>\frac{1}{2014}\) nên \(\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}< 0\)
=> x-2017=0
=>x=2017
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tìm x biết 6/(-3).(-5)+6/(-5).(-7)+...+6/(-2015).(-2017)=2014/2017.x
5 tháng 3 2023 lúc 9:38
Tìm x biết : \(\dfrac{x+4}{2014}\)+\(\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
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9 tháng 3 2018 lúc 20:39
Giải phương trình:
\(\dfrac{x-3}{2014}\)+\(\dfrac{x-2}{2015}\)=\(\dfrac{x-1}{1008}\)+\(\dfrac{x}{2017}\)-1
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
\(\Leftrightarrow\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{1008}-2+\dfrac{x}{2017}-1\) \(\Leftrightarrow\dfrac{x-3-2014}{2014}+\dfrac{x-2-2015}{2015}=\dfrac{x-1-2016}{1008}-\dfrac{x-2017}{2017}\) \(\Leftrightarrow\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{1008}+\dfrac{x-2017}{2017}\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0\)
Vì: \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\ne0\)
Suy ra: x -2017 = 0
=> x = 2017
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\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
⇔ \(\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{2008}-2+\dfrac{x}{2017}-1\)
⇔\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2008}+\dfrac{x-2017}{2017}\)
⇔\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2008}-\dfrac{x-2017}{2017}=0\)
⇔\(\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2008}-\dfrac{1}{2017}\right)=0\)
⇔x-2017=0
⇔x=2017
vậy phương trình có tập nghiệm là S={2017}
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\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
\(\Leftrightarrow\) \(\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{1008}-2+\dfrac{x}{2017}-1\)
\(\Leftrightarrow\) \(\dfrac{x-3}{2014}-\dfrac{2014}{2014}+\dfrac{x-2}{2015}-\dfrac{2015}{2015}=\dfrac{x-1}{1008}-\dfrac{2016}{1008}+\dfrac{x}{2017}-\dfrac{2017}{2017}\)
\(\Leftrightarrow\)\(\dfrac{x-3-2014}{2014}+\dfrac{x-2-2015}{2015}=\dfrac{x-1-2016}{1008}+\dfrac{x-2017}{2017}\)
\(\Leftrightarrow\)\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{1008}+\dfrac{x-2017}{2017}\)
\(\Leftrightarrow\) \(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{1008}-\dfrac{x-2017}{2017}=0\)
\(\Leftrightarrow\) \(\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow\) x - 2017 = 0
\(\Leftrightarrow\) x = 2017
Vậy.............
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