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5/6 + 11/12 + 19/20 + ... + 109/110
= (1 - 1/6) + (1 - 1/12) + (1 - 1/20) + ... + (1 - 1/110)
= (1 - 1/2×3) + (1 - 1/3×4) + (1 - 1/4×5) + ... + (1 - 1/10×11)
= (1 + 1 + 1 + ... + 1) - (1/2×3 + 1/3×4 + 1/4×5 + ... + 1/10×11)
( có 9 số 1)
= 1 x 9 - (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/10 - 1/11)
= 9 - (1/2 - 1/11)
= 9 - (11/22 - 2/22)
= 9 - 9/22
= 198/22 - 9/22
= 189/22
☆☆☆☆☆
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90
9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
9 – ( 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)
9 – (1 – 1/10) = 9 – 9/10 = 81/10
BÀI 5 :TÍNH NHAMH
1/2 + 5/6 +11/12 +19/20 + 29 / 30 + 41/42 + 55/56 + 71/72 +89/90
GIÚP MÌNH TÍNH NHÉ
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}\)
\(=\frac{81}{10}\)
Chú ý : dấu "." này là tương đương với dấu nhân nhé ( x )
Bài này dài quá nên khi mik làm nó cứ bị lệch dòng , thông cảm nhé : Trần Thanh Thảo
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
Cách 1 :
Ta có : \(1-\frac{4}{11}=\frac{7}{11};1-\frac{8}{15}=\frac{7}{15}\)
Vì \(\frac{7}{11}>\frac{7}{15}\)nên \(\frac{4}{11}< \frac{8}{15}\).
Cách 2 :
Vì \(\frac{4}{11}< \frac{4}{8}=\frac{1}{2};\frac{1}{2}=\frac{8}{16}< \frac{8}{15}\)nên \(\frac{4}{11}< \frac{8}{15}\).
~ HOK TỐT ~
đề thi vào trường chuyên nhe các bạn ai mà làm được mình sẽ k
dễ mà 1/2=1-1/2; 5/6=1-1/6; ... 89/90=1-1/90
Mà 1/2=1/1.2; 1/6=1/2.3; ... 1/90=1/9.10
Tương tự như thế sẽ tìm ra kq mik chỉ gợi ý thoy nhá !!!
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
= 1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90
= 9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
= 9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
= 9 – ( 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)
=9 – (1 – 1/10) = 9 – 9/10 = 81/10
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)=8+\frac{1}{10}=8,1\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+....+\left(1-\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9+\frac{1}{10}=9,1\)
\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)=\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)=\)
\(=9-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{9x10}\right)=\)
\(=9-\left(\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{10-9}{9x10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}=8\frac{1}{10}\)
= 1-1/6+1-1/12+1-1/20+.....+1-1/90
=(1+1+....+1)- (1/6+1/12+.....+1/90)
=1.14( 14 chữ số 1) - (1/2*3+1/3*4+....+1/9*10)
=14- (1/2-1/3+1/3-1/4+1/4-1/5+....+1/9-1/10) [ công thức là 1/2*3= 1/2-1/3]
= 14- (1/2-1/10)
= 14- 2/5
=68/5 (nhớ k cho mình nhé)
ủa sao cậu biết được rằng có 14 số vậy