(x-5)^x+2014=(x-5)^x+2015. tìm x
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=> (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
\(\frac{x+3}{2013}+1+\)\(\frac{x+4}{2012}+1+\frac{x+5}{2011}+1\)=\(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1+\frac{x}{2016}+1\)
\(\Rightarrow\frac{x+2016}{2013}+\frac{x+2016}{2012}+\frac{x+2016}{2011}=\frac{x+2016}{2014}+\frac{x+2016}{2016}\)
\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2016}=0\right)\)
Vì 1/2016+...+1/2011>0 nên (x+2016)=0
suy ra x= -2016
nếu đúng xin kết bạn
ch
\(Q=\frac{2014}{2015}\times3+\frac{2014}{2015}\times\frac{5}{6}-\frac{2014}{2015}\times\frac{10}{12}\)
\(Q=\frac{2014}{2015}\times\left(3+\frac{5}{6}-\frac{10}{12}\right)\)
\(Q=\frac{2014}{2015}\times3\)
\(Q=\frac{6042}{2015}\)
Vì \(\frac{x}{2}+\frac{x}{4}+\frac{x}{2014}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2015}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2014}\right)=x\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2015}\right)\)
Mà \(\frac{1}{2}+\frac{1}{4}+\frac{1}{2014}\ne\frac{1}{3}+\frac{1}{5}+\frac{1}{2015}\)
\(\Rightarrow x=0\)
\(\frac{3+5+7+...+2015}{2+4+6+...+2014}+x=1\)
\(\frac{\left(3+2015\right).\left[\left(2015-3\right):2+1\right]:2}{\left(2+2014\right).\left[\left(2014-2\right):2+1\right]:2}+x=1\)
\(\frac{1016063}{1015056}+x=1\)
\(x=1-\frac{1016063}{1015056}\)
\(x=-\frac{1}{1008}\)
Giải:
\(B=\left(5+10+15+...+2015\right)\times\left(20142014\times2015-20152015\times2014\right)\)
\(B=\left(5+10+15+...+2015\right)\times\left(10001\times2014\times2015-10001\times2015\times2014\right)\)
\(B=\left(5+10+15+...+2015\right)\times0\)
\(B=0\)
Chúc em học tốt!
B = (5+10+15+.........+2015) x (20142014 x 2015-20152015 x 2014)
B = (5+10+15+.........+2015) x ( 2014 x 1001 x 2015 - 2015 x 1001 x 2014 ] B = (5+10+15+.........+2015) x 0B = 0 nha
(x-5)^x+2014=(x-5)^x+2015
⇔ (x-5)^x-(x-5)^x=2015-2014
⇔ 0=1 (vô lý)
Vậy phương trình vô nghiệm.
(x-5)x+2015 - (x-5)x+2014 =0
(x-5)x+2014(x-5 -1) =0
+ x -5 =0 => x =5
+ x -6 =0 => x =6
Vậy x = 5 hoặc x =6
2 trường hợp nha