tinh nhanh 1/1x4 + 1/4x7 +1/7x10 +...+ 1/91x94
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Ta có :
1/1.4+1/4.7+...+1/91.94
=1/3.(1/1-1/4+...+1/91-1/94)
=1/3.(1/1-1/94)
=1/3.93/94
=31/94
Ta có :
1/1.4+1/4.7+...+1/91.94
=1/3.(1/1-1/4+...+1/91-1/94)
=1/3.(1/1-1/94)
=1/3.93/94
=31/94
1/1.4+1/4.7+1/7.10+...+1/91.94
=1/3.(3/1.4+3/4.7+3/7.10+...+3/91.94)
=1/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/91-1/94)
=1/3.(1-1-94)
=1/3.(93/94)
=31/94
Đặt A= 1/1*4+1/4*7+1/7*10+....+1/91*94
3A= 3/1*4+3/4*7+3/7*10+....+3/91*94
3A=1/1-1/4+1/4-1/7+1/7-1/10+............+1/91-1/94
3A=1-1/94=93/94=>A=93/94*1/3=31/94
=31/94 k mình nha bạn
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
\(S=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+...+\frac{1}{94\times97}+\frac{1}{97\times100}\)
\(S=\frac{1}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\times\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\times\frac{99}{100}\)
\(S=\frac{33}{100}\)
1/1.4+1/4.7+1/7.10+1/10.13+1/13.16
=1/3.(3/1.4+3/4.7+3/7.10+3/10.13+3/13.16)
=1/3.(1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16)
=1/3.(1/1-1/16)
=1/3.(16/16-1/16)=1/3.15/16=5/16
\(=\frac{1}{3}x\left(\frac{3}{1x4}+\frac{3}{4x7}+...+\frac{3}{77x80}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{80}\right)\)
\(=\frac{1}{3}\times\frac{79.}{80}\)
\(=\frac{79}{240}\)
Tk giúp mk nha cảm ơn !!
=1−14 +14 −110 +...+119 −122
=1−122
= \(\frac{21}{22}\)
k cho mình nha chắc chắn đúng 100 %
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}\)
\(=1-\frac{1}{22}\)
\(=\frac{21}{22}\)
tinh nhanh 1/1x4 + 1/4x7 +1/7x10 +...+ 1/91x94
Ta có :
1/1.4+1/4.7+...+1/91.94
=1/3.(1/1-1/4+...+1/91-1/94)
=1/3.(1/1-1/94)
=1/3.93/94
=31/94