Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(E=2\times4+4\times6+6\times8+...+98\times100\)
\(6\times E=2\times4\times6+4\times6\times\left(8-2\right)+6\times8\times\left(10-4\right)+...+98\times100\times\left(102-96\right)\)
\(=2\times4\times6+4\times6\times8-2\times4\times6+...+98\times100\times102-96\times98\times100\)
\(=98\times100\times102\)
\(\Rightarrow E=\frac{98\times100\times102}{6}=166600\)
\(A=2\times4+4\times6+6\times8+...+98\times100\)
\(6\times A=2\times4\times6+4\times6\times\left(8-2\right)+6\times8\times\left(10-4\right)+...+98\times100\times\left(102-96\right)\)
\(=2\times4\times6+4\times6\times8-2\times4\times6+6\times8\times10-4\times6\times8+...+98\times100\times102-96\times98\times100\)
\(=98\times100\times102\)
\(\Leftrightarrow A=\frac{98\times100\times102}{6}=166600\)
Câu hỏi tương tự nha
= 2 x ( 2 + 2 ) + 4 x ( 2 + 2 ) + 6 x ( 2 +2 ) +....+98 x ( 98 + 2 )
= 2 x 2 + 2 x 2 + 2 x 4 +4+......+98 x 98 = 2 x 98
= 2 x ( 2 + 4 + 6 +....+98 ) +( 2 x 2 + 4x4 + 6 x 6 +...+98 x 98 )
= 2 x 2450 + 40425 x 4
= 4900 + 161700 = 166600
Gọi biểu thức trên là A ta có:
Zô câu hỏi tương tự là cách giải
ĐS A = 49/200
\(B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{98.100}\)
\(\Rightarrow5B=\dfrac{20}{2.4}+\dfrac{20}{4.6}+...+\dfrac{20}{98.100}\)
\(\Rightarrow5B=10\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}\right)\)
\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(\Rightarrow5B=10.\dfrac{49}{100}\)
\(\Rightarrow5B=\dfrac{49}{10}\)
Vậy \(5B=\dfrac{49}{10}\)
Ta có: B = \(\dfrac{4}{2.4}\) + \(\dfrac{4}{4.6}\) + \(\dfrac{4}{6.8}\) + ... + \(\dfrac{4}{98.100}\).
=> \(\dfrac{B}{2}\) = \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) + \(\dfrac{2}{6.8}\) + ... + \(\dfrac{2}{98.100}\)
=\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\)
= \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\) = \(\dfrac{49}{100}\).
=> B = \(\dfrac{49}{200}\).
=> 5B = \(\dfrac{49}{200}\) . 5 = \(\dfrac{49}{40}\).
Vậy 5B = \(\dfrac{49}{40}\).
\(\frac{6}{2x4}+\frac{6}{4x6}+\frac{6}{6x8}+...+\frac{6}{98x100}\)
\(=3x\left(\frac{2}{2x4}+\frac{2}{4x6}+\frac{2}{6x8}+...+\frac{2}{98x100}\right)\)
\(=3x\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=3x\left(\frac{1}{2}-\frac{1}{100}\right)=3x\left(\frac{50}{100}-\frac{1}{100}\right)=3x\frac{49}{100}\)
\(=\frac{147}{100}\)
\(\frac{6}{2\cdot4}+\frac{6}{4\cdot6}+\frac{6}{6\cdot8}+...+\frac{6}{98\cdot100}\)
=\(\frac{3\cdot2}{2\cdot4}+\frac{3\cdot2}{4\cdot6}+\frac{3\cdot2}{6\cdot8}+...+\frac{3\cdot2}{98\cdot100}\)
=\(\text{}3\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{98\cdot100}\right)\)
=\(\text{}3\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
=\(\text{}3\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)
=\(\text{}3\cdot\frac{49}{100}=\frac{147}{100}\)
=1(2+1)+2(3+1)+3(4+1)+...+100(101+1)
=1.2+1+2.3+2+3.4+3+...+100.101+100
=(1.2+2.3+3.4+..+100.101)+(1+2+3+...+100)
=333300+5000
=338300
S=(2+98)*(4+6)+...+100+100+102
100*10+....+100+100*102
=224400