(1 + 1/2) x (1 + 1/3) x (1 + 1/4) x (1 + 1/x) = 1008 x 1/2
9.2^x.(2^1005-2^1002+.....+2^3-1)=2^1008-1 *
A=2^1005-2^1002+...+2^3-1 2^3 .
A=2^2.(2^1005-2^1002+......+2^3-1) 8.
A=2^1008-2^1005+.....+2^6-2^3
A=2^1005-2^1002+.....+2^3-1 8.A+A=2^1008-1 9.
A=2^1008-1
Thay 9.a=2^1008-1 vào * ta có: 2^x.(2^1008-1)=2^1008-1
2^x=(2^1008-1):(2^1008-1)
2^x=1
2^x=2^0
x=0
vậy x=0
Lập đề bài cho bài toán này
(1+1/2)(1+1/3)(1+1/4)….(1+1/x)=1008 1/2
=>\(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{x+1}{x}=1008.5\)
=>\(\dfrac{x+1}{2}=1008.5\)
=>x+1=2017
=>x=2016
Tìm x
(\(\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x...x\left(1+\frac{1}{x}\right)=1008\frac{1}{2}\)
\(A=\frac{2015+2016+2017}{2014+2015+2016+2017+2018}x1000\)
x - 1/1+2 - 1/1+2+3 - 1/1+2+3+4 - ...... - 1/1+2+3+......+2015=1/1008
\(\frac{1}{2}-\left(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+x}\right)=-\frac{503}{1008}\)tìm x
Rút gọn biểu thức A= 1 1/2 x 1 1/3 x 1 1/4 x ... x 1 1/2015 ta được A= ????????????????????
Các bạn cho mình hỏi luôn 1008 có phải là kết quả của bài ko ?????????????????????
GPT:
1)(x+1006/1000)+(x+1007/999)+(x+1008/998)+(x+1009/997)+(x+2022/4)=0
2)(x-999/99)+(x-896/101)+(x-769/103)=6
3)x4+x3+6x2=-5(x+1)
4)(x+1)(x+2)(X+3)(x+4)=24
5)(x+1)(x+2)(x+4)(x+5)=40
6)(x+1/x2+x+1)-(x-1/x2-x+1)=2(x+2)2/x6-1
7)2x4-7x3-2x2+13x+6/(x+1)(2x+1)=0
8)2x4-7x3-2x2+13x+6=0
Câu 2 sai đề nhé
Phải là:(x-999)/99+(x-896)/101+(x-789/103)=6
Tìm x, biết:
\(\frac{x-1}{2018}\)+ \(\frac{x-7}{503}\)= \(\frac{x-3}{1008}\)+ \(\frac{x-9}{670}\)
Chứng minh :
1/2 + 1/4 + 1/8 + ...+ 1/1024 < 1
1/2 +1/3 +1/4 +....+1/63 >2
\(\frac{x-1}{2018}+\frac{x-7}{503}=\frac{x-3}{1008}+\frac{x-9}{670}\)
\(\Leftrightarrow\frac{x-1}{2018}-1+\frac{x-7}{503}-4=\frac{x-3}{1008}-2+\frac{x-9}{670}-3\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{503}-\frac{1}{1008}-\frac{1}{670}\right)=0\)
\(\Rightarrow x=2019\)
#CBHT
Đặt A =1/2+1/4+1/8+...+1/1024
2A= 1+1/2+1/4+...+1/512
A= 1-1/1024
=>A<1hay ...
Cho \(a,b,x,y\) là các số thực thỏa mãn: \(x^2+y^2=1\) và \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}\) Chứng minh rằng: \(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{\left(a+b\right)^{1008}}\)