\(\frac{x+2016}{15}+\frac{x+2015}{14}=\frac{x+2014}{13}+\frac{x+2013}{12}\)
Những câu hỏi liên quan
a. \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
b. \(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)
a) x=-1
b) x=-2016
k mik nhé, ủng hộ nha:)
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Tìm số hữu tỉ x,biết:
a, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}+\frac{x+1}{15}\)
b, \(\frac{x+4}{2012}+\frac{x+3}{2013}+\frac{x+2}{2014}=\frac{x+1}{2015}+\frac{x-1}{2016}+\frac{x-2}{2017}\)
c, ( x - 2 )( x + 4 ) = 0
d, ( x + 1 )( x - 2 ) < 0
e, ( x - 2 )( x + \(\frac{1}{2}\)) > 0
giúp mình mới
Giải phương trình
\(\frac{x}{2-12}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
Phải là \(\frac{x}{2012}\)
\(\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x=2012\)
Vậy ...
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Tìm x thuộc Q biết
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}+\frac{x+1}{13}+\frac{x+1}{14}\)
b) \(\frac{x+4}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}+\frac{x+5}{2015}\)
tìm x : \(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
=> \(\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+3}{2014}+1\right)+\left(\frac{x+4}{2013}+1\right)\)
=> \(\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)
=> (x + 2017)(1/2015 + 1/2016 - 1/2014 - 1/2013) = 0
=> x + 2017 = 0
=> x = -2017
\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
\(\Leftrightarrow\frac{x+2}{2015}+1+\frac{x+1}{2016}+1=\frac{x+3}{2014}+1+\frac{x+4}{2013}+1\)
\(\Leftrightarrow\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Dễ thấy cái ngoặc to < 0
=> x=-2017
Tìm x: \(\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016}\)
\(\Leftrightarrow\left(\frac{x+4}{2013}+1\right)+\left(\frac{x+3}{2014}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)
\(\Rightarrow x+2017=0\Rightarrow x=-2017\)
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Giải phương trình: \(\frac{x+4037}{2016}+\frac{x+6050}{2015}+\frac{x+8061}{2014}=\frac{x+18122}{2013}\)
bài này mình lm ra rrrrrrrrr, khỏi giải nha mấy bn
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Tìm x biết : \(\frac{x+2}{2013}+\frac{x+1}{2014}=\frac{x}{2015}+\frac{x-1}{2016}\)
\(\frac{x+2}{2013}+\frac{x+1}{2014}=\frac{x}{2015}+\frac{x-1}{2016}\)
\(\Leftrightarrow\)\(\frac{x+2}{2013}+1+\frac{x+1}{2014}+1=\frac{x}{2015}+1+\frac{x-1}{2016}+1\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2014}=\frac{x+2015}{2015}+\frac{x+2015}{2016}\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
Do\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}>0\)
=>x+2015=0
<=>x=-2015
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=> \(\frac{x+2015-2013}{2013}+\frac{x+2015-2014}{2014}=\frac{x+2015-2015}{2015}+\frac{x+2015-2016}{2016}\)
<=> \(\frac{x+2015}{2013}-1+\frac{x+2015}{2014}-1=\frac{x+2015}{2015}-1+\frac{x+2015}{2016}-1\)
<=> \(\frac{x+2015}{2013}+\frac{x+2015}{2014}-\frac{x+2015}{2015}-\frac{x+2015}{2016}=0\)
<=> \(\left(x+2015\right).\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
<=> x + 2015 = 0 Vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)
<=> x = -2015
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\(\frac{x+1}{2013}+\frac{x+1}{2014}+\frac{x+1}{2015}=\frac{x+1}{2016}+\frac{x+1}{2017}\)
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