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uh.cậu là fan của bts hả.mình cũng thế,nhưng mình thích red velvet hơn
Biến đổi VT ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=VP\)
\(\Rightarrowđpcm\)
1/1+2 + 1/+1+2+3 + ... + 1/1+2+3+...+2014
= 1/(1+2).2:2 + 1/(1+3).3:2 + ... + 1/(1 + 2014).2014:2
= 2/2.3 + 2/3.4 + ... + 2/2014.2015
= 2.(1/2.3 + 1/3.4 + ... + 1/2014.2015)
= 2.(1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2014 - 1/2015)
= 2.(1/2 - 1/2015)
= 2.1/2 - 2.1/2015
= 1 - 2/2015
= 2013/2015
1/1+2 + 1/+1+2+3 + ... + 1/1+2+3+...+2014
= 1/(1+2).2:2 + 1/(1+3).3:2 + ... + 1/(1 + 2014).2014:2
= 2/2.3 + 2/3.4 + ... + 2/2014.2015
= 2.(1/2.3 + 1/3.4 + ... + 1/2014.2015)
= 2.(1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2014 - 1/2015)
= 2.(1/2 - 1/2015)
= 2.1/2 - 2.1/2015
= 1 - 2/2015
= 2013/2015
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
\(A=\frac{2^{100}-1}{2^{100}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
hok tốt!!
1,\(\left(\frac{7}{2}-2x\right).\frac{4}{3}=\frac{22}{3}\)
\(x.\left(\frac{7}{2}-2\right)=\frac{22}{3}:\frac{4}{3}=\frac{22}{3}.\frac{3}{4}=\frac{11}{2}\)
\(x.\frac{3}{2}=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{3}{2}=\frac{11}{2}.\frac{2}{3}=\frac{11}{3}\)
Ta có \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}>0\)
=> \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}>1\) (1)
Ta lại có \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
< \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
< \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
< \(1-\frac{1}{100}< 1\)
=> \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)\(< 1+1\)
=> \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)\(< 2\) (2)
Từ (1) và (2) => \(1< 1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 2\)
=> \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)không là số tự nhiên
\(1)\frac{1}{2}x-\frac{3}{5}=\frac{-4}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-1}{5}\)
\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}\)
\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)
\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)
Đến đây tìm được rồi nhé
3,4, áp dụng bài 1,2 rồi làm :v
Đặt \(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
\(\Rightarrow S< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow S< 1-\frac{1}{100}< 1\Rightarrow S< 1\)
Làm vui đó chủ yếu là nghe link gửi
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}< 1\)
\(A< 1\left(đpcm\right)\)