giải phương trình
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{20121}\)
K
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11 tháng 1 2017
Theo bài ra , ta có :
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)=\left(\frac{x+3}{2013}+1\right)+\left(\frac{x+4}{2012}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+2014}{2014}\right)+\left(\frac{x+1+2015}{2015}\right)=\left(\frac{x+3+2013}{2013}\right)+\left(\frac{x+4+2012}{2012}\right)\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Vì \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)>0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
Vậy \(x=-2016\)
Tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)
Chúc bạn học tốt =))
11 tháng 1 2017
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\frac{x+2}{2014}+1+\frac{x+1}{2015}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\frac{x+2+2014}{2014}+\frac{x+1+2015}{2015}=\frac{x+3+2013}{2013}+\frac{x+4+2012}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\left(x+2016\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
MÀ \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
KN
11 tháng 2 2020
a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)
\(\Rightarrow x+10=0\Rightarrow x=-10\)
Vậy x = -10
b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)
\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)
\(\Rightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}\)\(+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Mà \(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)nên x - 2012 = 0
Vậy x = 2012
11 tháng 2 2020
a, (x+1)/9 +1 + (x+2)/8 = (x+3)/7 + 1 + (x+4)/6 + 1
<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6
<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0
vì 1/9 +1/8 -1/7 - 1/6 khác 0
=> x+10=0
=> x=-10
1 tháng 9 2019
Đặt \(\sqrt{x-2013}=a\left(a>0\right)\)
\(\sqrt{y-2014}=b\left(b>0\right)\)
\(\sqrt{z-2015}=c\left(c>0\right)\)
Có \(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
<=> \(\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)
<=> \(\frac{4a-4-a^2}{4.a^2}+\frac{4b-4-b^2}{4b^2}+\frac{4c-4+c^2}{4c^2}=0\)
<=>\(\frac{-\left(a^2-4a+4\right)}{4a^2}-\frac{b^2-4b+4}{4b^2}-\frac{c^2-4c+4}{4c^2}=0\)
<=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}=0\).
Có \(\frac{\left(a-2\right)^2}{4a^2}\ge0\forall a>0\)
\(\frac{\left(b-2\right)^2}{4b^2}\ge0\forall b>0\)
\(\frac{\left(c-2\right)^2}{4c^2}\ge0\forall c>0\)
=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}\ge0\) với moi a,b,c >0
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}\sqrt{x-2013}=2\\\sqrt{y-2014}=2\\\sqrt{z-2015}=2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x-2013=4\\y-2014=4\\z-2015=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)(t/m)
Vậy \(\left(x,y,z\right)\in\left\{\left(2017,2018,2019\right)\right\}\)
N
17 tháng 3 2019
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 ...
CN
0
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\) <=> \(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\) <=> \(\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\) <=> \(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\) <=> \(x+2016\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\) <=> x + 2016 = 0 <=> x = -2016