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Ta có:
\(\Rightarrow A=B.\)
\(\Rightarrow A^{2017}=B^{2017}\)
\(\Rightarrow\left(A^{2017}-B^{2017}\right)^{2018}=\left(B^{2017}-B^{2017}\right)^{2018}=0^{2018}=0.\)
Vậy \(\left(A^{2017}-B^{2017}\right)^{2018}=0.\)
Chúc bạn học tốt!
ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1
=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2
=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019
=> B= 2019 *(1/2+1/3+...+1/2019)
=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)
=> A/B= 1/2019
a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)
b, c cùng 1 câu phải k
ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)
A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)
NHA
HỌC TỐT
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2017}\left(1+2+...+2017\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{2017}.\frac{2017\left(2017+1\right)}{2}\)
\(=1+\frac{2.3}{2.2}+\frac{3.4}{3.2}+....+\frac{2017.2018}{2017.2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2018}{2}\)
\(=\frac{2+3+4+...+2018}{2}\)
\(=\frac{\frac{2018\left(2018+1\right)}{2}-1}{2}\)
\(=1018585\)
Nguyễn Tiến Đạt
a)\(|3x-5|=|x+2|\)
=> Ta có 2 trường hợp
*) TH1: 3x-5=x+2
=>3x-x=2+5
=>2x=7
=>x=7:2\(\Rightarrow x=\frac{7}{2}\)
*)TH2: -3x+5=x+2
\(\Rightarrow5-3x=x+2\)
\(\Rightarrow5-2=x+3x\)
\(\Rightarrow3=4x\)
\(\Rightarrow x=3:4\Rightarrow x=\frac{3}{4}\)
Vậy \(x\in\left\{\frac{7}{2};\frac{3}{4}\right\}\)