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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+.....+\frac{2}{73.76}\)
\(=\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{73.76}\right)\)
\(=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{73}-\frac{1}{76}\right)\)
\(=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(=\frac{2}{3}.\frac{9}{38}=\frac{3}{19}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(A=\frac{2}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(A=\frac{33}{50}\)
Vậy \(A=\frac{33}{50}\)
Chúc bạn học tốt ~
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{23}{12}\)
\(\frac{314}{105}\)
\(\frac{59}{60}\)
\(\frac{199}{90}\)
\(\frac{1}{18}\)
\(\frac{13}{36}\)
\(\frac{4}{221}\)
\(\frac{4}{85}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\)\(1\frac{2}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+5\frac{3}{7}\)
\(=\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{3}\cdot\frac{3}{2}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(=\frac{4}{2}+\frac{38}{7}\)
\(=2+\frac{38}{7}\)
\(=\frac{14}{7}+\frac{38}{7}\)
\(=\frac{52}{7}\)
\(b.1\frac{1}{3}-1\frac{1}{4}:1\frac{1}{2}+2\frac{3}{4}\cdot3\frac{2}{3}\)
\(=\frac{4}{3}-\frac{5}{4}:\frac{3}{2}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{4}\cdot\frac{2}{3}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{6}+\frac{121}{12}\)
\(=\frac{16}{12}-\frac{10}{12}+\frac{121}{12}\)
\(=\frac{6}{12}+\frac{121}{12}\)
\(=\frac{127}{12}\)
\(c.7\cdot\frac{2}{3}-\frac{2}{5}:\frac{1}{2}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{2}{5}\cdot\frac{2}{1}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{14}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{70}{15}-\frac{12}{15}-\frac{10}{15}\)
\(=\frac{58}{15}-\frac{10}{15}\)
\(=\frac{48}{15}=\frac{16}{5}\)
\(\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(2+\frac{38}{7}\)
\(\frac{52}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(8\frac{7}{10}+2\frac{3}{4}=\frac{87}{10}+\frac{11}{4}=\frac{174}{20}+\frac{55}{20}=\frac{229}{20}\)
Bạn chỉ cần đưa về phân số xong tính bình thường. Muốn đổi từ hỗn số sang phân số, ta chỉ cần lấy phần nguyên nhân cho mẫu rồi cộng với tử là xong. Chứ bạn cứ hỏi mấy bài dễ như thế này thì k giỏi đc đâu!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
ở dãy 1 thì số đứng sau bằng tổng hai số đứng trước
ta có 5 số tiếp theo la 40,74, 136,...
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 6/9+5/7+1/3=2/3+5/7+1/3=5/7+1=12/7
b, 17/7+6/5-20/14=17/7+6/5-10/7=6/5+1=11/5
c,2/5x1/4+3/4x2/5=2/5x(1/4+3/4)=2/5x1=2/5
d, 6/11:4/6+5/11:2/3=6/11:2/3+5/11:2/3=(6/11+5/11):2/3=3/2
nha
C = 3/4.7 + 3/7.10 + 3/10.13 + ... + 3/73.76
=1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ... + 1/73 - 1/76
=1/4 - 1/76
=18/76
\(C=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+......+\frac{1}{73}-\frac{1}{76}\)
\(=\frac{1}{4}-\frac{1}{76}\)
\(=\frac{19}{76}-\frac{1}{76}\)
\(=\frac{18}{76}=\frac{9}{38}\)