1/2+1/6+1/12+1/20+1/30+1/42+1/56 +1/72+1/90+? các bạn gi cách giải cho mình nhé.mình k cho
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\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}\)
\(=\frac{-79}{90}\)
1/90 - 1/72 - 1/56 - ... - 1/6 - 1/2
= 1/90 - (1/2 + 1/6 + ... + 1/56 + 1/72)
= 1/90 - (1/1×2 + 1/2×3 + ... + 1/7×8 + 1/8×9)
= 1/90 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8 + 1/8 - 1/9)
= 1/90 - (1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
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\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}=\frac{9}{10}\)
k cho mk nha
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}\)
\(\frac{9}{10}\)
ta có: 1/2=1/1x2
1/6=1/2x3
1/12=1/3x4
1/20=1/4x5
1/30=1/5x6
1/42=1/6x7
1/56=1/7x8
1/72=1/8x9
1/90=1/9x10
1/110=1/10x11
tiếp theo bn tiếp tục nhé
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110
=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x1
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11
=1-1/11
=10/11
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{11\cdot12}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{3}-\frac{1}{12}=\frac{4}{12}-\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
Chú ý: \(\cdot=\times\)
Đặt \(A=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{12}=\frac{1}{4}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
em lớp 6 nha
B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72
B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
B=1+0-0-0-0-0-0-0-1/9
B=1-1/9
B=8/9
k em nha
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\) \(+?\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\) \(+?\)
\(=>?=\frac{1}{10\cdot11}=\frac{1}{110}\)
Vậy \(?\) là \(\frac{1}{110}\)
= 0,9 nhé
CHÚC BẠN HỌC GIỎI
TK MÌNH NHÉ