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(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!
\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}+\frac{x+4}{2011}=0\)
\(\Leftrightarrow\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)+\left(\frac{x+4}{2011}+1\right)=0+1+1+1+1\)
\(\Leftrightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}+\frac{x+2015}{2011}=4\)
\(\Leftrightarrow\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=4\)
\(\Leftrightarrow x+2015=4:\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)\)
\(\Leftrightarrow x+2015=2012,499379\)
\(\Leftrightarrow x=2012,499379-2015\)
\(\Leftrightarrow x=-2,5006118\)
P/s : Số xấu quá !
b) Ta có: \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2\)
\(=\left(ab^2-cb^2\right)+\left(ca^2-c^2a\right)+\left(bc^2-ba^2\right)\)
\(=b^2\left(a-c\right)+ca\left(a-c\right)+b\left(c^2-a^2\right)\)
\(=\left(a-c\right)\left(b^2+ca\right)-b\left(a-c\right)\left(a+c\right)\)
\(=\left(a-c\right)\left(b^2+ca-ba-bc\right)\)
\(=\left(a-c\right)\left[b\left(b-a\right)+c\left(a-b\right)\right]\)
\(=\left(a-c\right)\left[b\left(b-a\right)-c\left(b-a\right)\right]\)
\(=\left(a-c\right)\left(b-a\right)\left(b-c\right)\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
=>\(2\left(ab+bc+ac\right)=0\)
=>ab+bc+ac=0
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
=>\(\dfrac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{3}{abc}\)
=>\(\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)
\(\Leftrightarrow\left(ab+bc\right)^3-3\cdot ab\cdot bc\cdot\left(ab+bc\right)+\left(ac\right)^3=3\left(abc\right)^2\)
=>\(\left(-ac\right)^3-3\cdot ab\cdot bc\cdot\left(-ac\right)+\left(ac\right)^3-3\left(abc\right)^2=0\)
=>\(-a^3c^3+a^3c^3+3a^2b^2c^2-3a^2b^2c^2=0\)
=>0=0(đúng)
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 ~
1)
\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}+\dfrac{x-3}{2012}+...+\dfrac{x-2014}{1}=2014\)
\(\Leftrightarrow\left(\dfrac{x-1}{2014}-1\right)+\left(\dfrac{x-2}{2013}-1\right)+...+\left(\dfrac{x-2014}{1}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}+...+\dfrac{x-2015}{1}=0\)
\(\Leftrightarrow\left(x-2025\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1}\right)=0\)
\(\Leftrightarrow x=2015\)
Vậy \(S=\left\{2015\right\}\)