1.Tính
(1-1/2)*(1-1/3)*(1-1/4)*(1-1/5)*...*(1-1/2017)*(1-1/2018)
gúp mk vs nha mk đg cần gấp
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BN
0
NT
35
31 tháng 12 2018
1 + 2 -3-4 + 5 + 6-7-8+...+2017+2018
= 1 + (2-3-4+5) + (6-7-8+9) + ...+ (2014-2015-2016+2017) + 2018
= 1 + 0+0+...+0+2018
=2 019
22 tháng 4 2018
Ta có :
\(A=\frac{2018^{2017}+1}{2018^{2017}-1}=\frac{2018^{2017}-1+2}{2018^{2017}-1}=\frac{2018^{2017}-1}{2018^{2017}-1}+\frac{2}{2018^{2017}-1}=1+\frac{2}{2018^{2017}-1}\)
\(B=\frac{2018^{2017}-1}{2018^{2017}-3}=\frac{2018^{2017}-3+2}{2018^{2017}-3}=\frac{2018^{2017}-3}{2018^{2017}-3}+\frac{2}{2018^{2017}-3}=1+\frac{2}{2018^{2017}-3}\)
Vì \(2018^{2017}-1>2018^{2017}-3\) nên \(\frac{2}{2018^{2017}-1}< \frac{2}{2018^{2017}-3}\)
\(\Rightarrow\)\(1+\frac{2}{2018^{2017}-1}< 1+\frac{2}{2018^{2017}-3}\)
\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
22 tháng 4 2018
ta có nếu \(\frac{a}{b}\)>1 thì \(\frac{a}{b}\)>\(\frac{a+m}{b+m}\)
mà B> nên B=\(\frac{2018^{2017}-1}{2018^{2017}-3}\)>\(\frac{2018^{2017}-1+2}{2018^{2017}-3+2}\)=\(\frac{2018^{2017}+1}{2018^{2017}-1}\)=A
vậy B>A
VM
13 tháng 11 2019
b) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\Rightarrow\frac{3}{5}x=\left(-\frac{1}{7}\right)+\frac{1}{2}\)
\(\Rightarrow\frac{3}{5}x=\frac{5}{14}\)
\(\Rightarrow x=\frac{5}{14}:\frac{3}{5}\)
\(\Rightarrow x=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}.\)
c) \(5-\left|3x-1\right|=3\)
\(\Rightarrow\left|3x-1\right|=5-3\)
\(\Rightarrow\left|3x-1\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3:3\\x=\left(-1\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{1;-\frac{1}{3}\right\}.\)
d) \(\left(1-2x\right)^2=9\)
\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow1-2x=\pm3.\)
\(\Rightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-2\right):2\\x=4:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{-1;2\right\}.\)
Chúc bạn học tốt!
NH
0
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot4\cdot....\cdot2016\cdot2017}{2\cdot3\cdot4\cdot5\cdot....\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
1/2018