giải bất phương trình:
x+2015/2 + x+2014/3 > -x-2013/-4 + -x-2012/-5
giải phương trình :
(x/2012)+(x+1/2013)+(x+2/2014)+(x+3/2015)+(x+4/2016)=5
pt <=> (x/2012 - 1) + (x+1/2013 - 1) + (x+2/2014 - 1) + (x+3/2015 - 1) + (x+4/2016 - 1) = 0
<=> x-2012/2012 + x-2012/2013 + x-2012/2014 + x-2012/2015 + x-2012/2016 = 0
<=> (x-2012).(1/2012+1/2013+1/2014+1/2015+1/2016) = 0
<=> x-2012 = 0 ( vì 1/2012+1/2013+1/2014+1/2015+1/2016 > 0 )
<=> x=2012
Vậy x=2012
Tk mk nha
Ta có :
\(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Leftrightarrow\)\(\left(\frac{x}{2012}-1\right)+\left(\frac{x+1}{2013}-1\right)+\left(\frac{x+2}{2014}-1\right)+\left(\frac{x+3}{2015}-1\right)+\left(\frac{x+4}{2016}-1\right)=5-5\)
\(\Leftrightarrow\)\(\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Leftrightarrow\)\(\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\ne0\)
\(\Rightarrow\)\(x-2012=0\)
\(\Rightarrow\)\(x=2012\)
Vậy \(x=2012\)
Chúc bạn học tốt ~
giải phương trình sau (x-2) / 2012 + (x-4) / 2014 + (x-6) / 2016 = (x-1) / 2011+ (x-3 ) /2013+ (x-5) /2015
Giải phương trình và bất phương trình : a)(x-1)/2015+x/2014+2/1006=(x-3)/2013+x/2012+1/1007 b) 2y^2 -5 / y^3-1 ≥4/1+y+y^2 +1/1-y
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\) giải phương trình
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 =))
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\frac{x+2}{2014}+1+\frac{x+1}{2015}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\frac{x+2+2014}{2014}+\frac{x+1+2015}{2015}=\frac{x+3+2013}{2013}+\frac{x+4+2012}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\left(x+2016\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
MÀ \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
Giải phương trình :\(\frac{x+1}{2015}+\frac{x+2}{2014}+\frac{x+3}{2013}+\frac{x+4}{2012}+\frac{x+5}{2011}+\frac{x+6}{2010}=0\)
Giải phương trình:
\(\frac{x-4}{2012}+\frac{x-3}{2013}=\frac{x-2}{2014}+\frac{x-1}{2015}\)
cho mk hỏi cách giải bài đó đi đáp án mk pk rồi
Giải các phương trình và bất phương trình sau
a) \(\frac{x-1}{2015}+\frac{x}{2014}+\frac{2}{1006}=\)\(\frac{x-3}{2013}+\frac{x}{2012}+\frac{1}{1007}\)
b) \(\frac{4}{1+y+y^2}+\frac{1}{1-y}\le\frac{2y^2-5}{y^3-1}\)
Giải phương trình: (x-1)/2015+x/2014+2/1006=(x-3)/2013+x/2012+1/1007
Nhanh lên giúp mik với mn ơi
(x-1)/2015 + x/2014 + 1/503 - (x-3)/2013 - x/2012 - 1/1007 =0
(x-2016)/2015 + (x-2016)/2014 - (x-2016)/2012 - (x-2016)/2013 = 0
(x-2016) ( 1/2015 + 1/2016 - 1/2013 - 1/2012) = 0
Mà 1/2015 + 1/2016 - 1/2013 - 1/2012 khác 0
Suy ra x -2016=0
x=2016
Chỗ nào thắc mắc nhớ hỏi mik nhe!
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giải pt x-5/2012+x-4/2013=x-3/2014+x-2/2015 giải hộ với
\(\dfrac{x-5}{2012}+\dfrac{x-4}{2013}=\dfrac{x-3}{2014}+\dfrac{x-2}{2015}\)
\(\Rightarrow\left(\dfrac{x-5}{2012}-1\right)+\left(\dfrac{x-4}{2013}-1\right)=\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}=\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}\)
\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}-\dfrac{x-2017}{2015}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)
\(\Rightarrow x-2017=0\Leftrightarrow x=2017\)
Vậy x = 2017