Tính nhanh: \(89^2+11^2+22.89\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
892+112+89.22
=892+112+89.11+11.89.
=89.(11+89)+11.(11+89)
=89.100+11.100
=8900+1100
=10000
Chúc em học tốt^^
892 + 112 + 89.22
= (892 + 89.11) + (112 + 89.11)
= 89.(89 + 11) + 11.(11 + 89)
= 89.100 + 11.100
= 100.(89 + 11)
= 100.100
= 10000
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2=1-1/2
5/6=1-1/6
11/12=1-1/12
109/110=1-1/110
A=10-(1/2+1/6+...+1/110)
B=(1/2+1/6+...+1/110)=[(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/9-1/10)+(1/10+1/11)
B=1+1/11
A=10-(1+1/11)=10+1/11
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+1+1+1+1+1+1+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ở đây bạn thiếu phân số 29/30 rồi, mk thêm vào nhé
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{5}{6}+\frac{1}{6}\right)+\left(\frac{11}{12}+\frac{1}{12}\right)+...+\left(\frac{89}{90}+\frac{1}{90}\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(=1+1+1+...+1-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{89}-\frac{1}{90}\right)\)
( 9 số 1 )
\(=9-\left(\frac{1}{1}-\frac{1}{90}\right)=9-\frac{89}{90}=8\frac{1}{90}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
A=1/2+ 5/6 + 11/12 + 19/20 + 29 30 + 41/42 + 55/56 + 71/72 + 89/90
![](https://rs.olm.vn/images/avt/0.png?1311)
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 !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
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
Đề sai phải đổi \(\frac{11}{16}\) thành \(\frac{11}{12}\)
\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=9-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10-9}{9.10}\right)\)
\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=8\frac{1}{10}\)
892+112+22.89
=892+2.89..11+112
=(89+11)2
=1002
=10000
Tuy mình chưa học lớp 8 nhưng mình cũng có thể làm được bài này.
= 892 + 2.89.11 + 112 = (89 +11)2 = 1002 = 10000