ahihi

\(A=3+3^2+...+3^{50}\)

\(\Rightarrow3A=3^2+3^3+...+3^{50}+3^{51}\)

\(\Rightarrow3A-A=3^{51}-3\)

\(\Rightarrow2A=3^{51}-3\)

\(\Rightarrow A=\frac{3^{51}-3}{2}\)

\(B=2-2^2+2^3-2^4+...+2^{2019}-2^{2020}\)

\(2B=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)

\(B+2B=2-2^{2021}\)

\(3B=2-2^{2021}\)

\(B=\frac{2-2^{2021}}{3}\)

\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2008.2009}\)

\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}\)

\(C=1-\frac{1}{2009}\)

\(C=\frac{2008}{2009}\)

\(D=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(D=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(D=\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(D=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

Ta có : \(A=3+3^2+3^3+.....+3^{2016}\)

\(\Rightarrow3A=3^2+3^3+3^4+......+3^{2017}\)

\(\Rightarrow3A-A=3^{2017}-3\)

\(\Rightarrow2A=3^{2017}-3\)

\(\Rightarrow A=\frac{3^{2017}-3}{2}\)

\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{1024}\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{512}\)

\(\Rightarrow2B-B=1-\frac{1}{1024}\)

\(\Rightarrow B=\frac{1023}{1024}\)

bài 1:

<=> \(x\left(x^2-\frac{9}{16}\right)=0\)

TH1:x=0

TH2: \(x^2-\frac{9}{16}=0\)

=> \(x^2=\frac{9}{16}\)

TH2a: \(\Rightarrow x=\frac34\)

\(TH2b:x=-\frac34\)

bài 2:

1) <=> \(2N=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\cdots+\frac{2}{98\cdot99\cdot100}\)

Áp dụng công thức: \(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+1\right)}\) ta có:

\(2N=\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+\cdots+\left(\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)

\(2N=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\)

\(2N=\frac{4949}{9900}\)

\(\Rightarrow N=\frac{4949}{19800}\)

2) <=> \(5N=5^2+5^3+5^4+\cdots+5^{101}\)

=> \(5N-N=\left(5^2+5^3+5^4+\cdots+5^{101}\right)-\left(5+5^2+5^3+\cdots+5^{99}+5^{100}\right)\)

\(4N=5^{101}-5\)

=> \(N=\frac{\left(5^{101}-5\right)}{4}\)

em không biết 

Câu hỏi của ✨♔♕ Saiko ♕♔✨ - Toán lớp 6 - Học toán với OnlineMath

các bạn giải nhanh nha mai mk nộp bài nhanh nha

khó quá bn ơi

2b,