Tìm x biết : (x+1)+(x+2)+....+(x+2015)=2015.2016
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Tìm: \(\frac{x}{2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2015.2016}=\frac{2015}{4032}\)
\(\frac{x}{2}+\frac{x}{2.3}+\frac{x}{3.4}+.....+\frac{x}{2015.2016}=\frac{2015}{4032}\)
\(x.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2015.2016}\right)=\frac{2015}{4032}\)
\(x.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)=\frac{2015}{4032}\)
\(x.\left(1-\frac{1}{2016}\right)=\frac{2015}{4032}\)
\(x.\frac{2015}{2016}=\frac{2014}{4032}\)
\(x=\frac{2015}{4032}:\frac{2015}{2016}\)
\(x=\frac{1}{2}\)
\(=\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{2015}-\frac{x}{2016}=\frac{2015}{4023}\)
\(=\frac{x}{1}-\frac{x}{2016}=\frac{2015}{4023}\)
\(=\frac{2015}{2016}x=\frac{2015}{4023}\)
=> x = \(\frac{2015}{4023}\cdot\frac{2016}{2015}\)= 2016/4023
Tìm x
\(\frac{x-2017}{2015.2016}+\frac{x-2018}{2016.2017}+\frac{x-2019}{2017.2018}+\frac{x-2020}{2018.1019}=\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{1018}\)
Tìm x biết x+2x+3x+...+2014x=2015.2016
1:Tìm x, biết:
a) (3/5)^2x-1 = 125/27
b)1/2+(1/2.3)+(1/3.4)+...+(1/2015.2016)-2x=1/2-1/2016
c)/2x^2+/x-2//=2x^2+3
Tìm x, biết:
a, (3/5)^2x-1=125/27
b,1/2+1/2.3+1/3.4+...+1/2015.2016 - 2x=1/2-1/2016
c,/2x^2+/x-2//=2x^2+3
Tìm X biết : B= x+(x+1)+(x+2)+(x+3)+.....+2015=2015
Tìm x biết : 2x+4x+6x+....+2014x=2015.2016
2x+4x+6x+...+2014x=2015.2016
=>(2+4+6+...+2014).x=2015.2016
tổng trong ngoặc có:(2014-2):2+1=1007(số hạng)
=>tổng= (2014+2).1007:2=1015056
=>1015056.x=4062240
=>x=4030/1007
Tìm x thuộc Z biết:
1) 2016+2015+2014+...+x = 2016
2) 1+2+3+...+x = 1275
3) | x+2015 | + | x+2016| = 1
thiện xạ 5a3 có thể giải chi tiết ra đc k? Mk cần cách lm
2) 1+2+3+...+x=1275
Có SSH là: (x+1):1+1=x(SH)
=> (x+1).x:2=1275
=>(x+1).x=1275.2
=>(x+1).x=2550
=>(x+1).x=51.50
=>x=50
3) |x+2015|+|x+2016|=1
Ta thấy |x+2015| và |x+2016| > hoặc = 0 với mọi x
=> 1= 0+1=1+0
+) x+2015=0=>x=-2015
x+2016=1=>x=-2015
+) x+2015=1=>x=-2014
x+2016=0=> x=-2016
Vậy xE{...}
Tìm x,biết:
x+2015/5 + x+2014/6 = x+2017/3 + x+2018/2
Hướng dẫn: x+2015/5+1 + x+2014/6+1 = x+2017/3+1 + x+2018/2+1
=> (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