\(\frac{1}{1x4}\)+\(\frac{1}{2x6}\)+\(\frac{1}{3x8}\)+...........+\(\frac{1}{99x200}\)=?
1/1x4+1/2x6+1/3x8+.....+1/99x200 = ?
1/(1x4)+1/(2x6)+1/(3x8)+.....+1/(2017x4036)
Chứng minh rằng : A=3/1x4 + 3/2x6 + 3/3x8 + ..... +1/2012x1342 < 1,5
Tính :
A = \(\frac{5}{3x8}+\frac{8}{8x16}+\frac{1}{16x3}+\frac{1}{24x7}+\frac{1}{28x8}+\frac{1}{32x5}\)
Tính nhanh:
\(\frac{1}{1x4}+\frac{1}{4x7}+\frac{1}{7x10}+...+\frac{1}{97x100}\)
1/1*4 + 1/4*7 + 1/7*10 + ... + 1/97*100
= 1/3(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/97*100)
= 1/3(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100)
= 1/3(1 - 1/100)
= 1/3*99/100
= 33/100
trả lời
=33/100
chúc bn
học tốt
Chứng minh rằng:A=3/(1x4)+3/(2x6)+3/(3x8)+...+1/(2012x2013)<1,5
Nếu sai đề thì thông cảm nha! Đề nó ghi vậy.
A=\(\frac{1}{1x2}+\frac{1}{1x3}+\frac{1}{1x4}+\frac{1}{1x5}+\frac{12}{10}\)
Tìm giá trị của A
\(\frac{149}{60}\)
k mk nha mk đag âm
\(\frac{1}{1\times2}\) + \(\frac{1}{1\times3}\) + \(\frac{1}{1\times4}\) + \(\frac{1}{1\times5}\) + \(\frac{12}{10}\)
= \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{12}{10}\)
= \(\frac{149}{60}\)
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{12}{10}\)
\(A=\frac{30}{60}+\frac{20}{60}+\frac{15}{60}+\frac{12}{60}+\frac{72}{60}\)
\(A=\frac{30+20+15+12+72}{60}\)
\(A=\frac{149}{60}\)
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\(\frac{1}{1x4}+\frac{1}{4x7}+\frac{1}{7x10}+.......+\frac{1}{77x80}\)
giúp mình nha
Ai nhanh mình sẽ tick
\(=\frac{1}{3}x\left(\frac{3}{1x4}+\frac{3}{4x7}+...+\frac{3}{77x80}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{80}\right)\)
\(=\frac{1}{3}\times\frac{79.}{80}\)
\(=\frac{79}{240}\)
Tk giúp mk nha cảm ơn !!
Cho: S=\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+...+\frac{3}{100x103}\). Chứng minh S<1
S=1/1-1/4+1/4+1/7-1/7+1/10+...+1/100-1/103
S=1/1-1/103
S=102/103
Vì 102/103<1 nên S<1
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{100\cdot103}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)
\(S=1-\frac{1}{103}\)
\(S=\frac{102}{103}< 1\)
\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+.......+\frac{3}{100x103}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}....+\frac{1}{100}-\frac{1}{103}\)
\(=\frac{1}{1}-\frac{1}{103}\)
=\(\frac{102}{103}\)