\(\frac{1}{9}\)+\(\frac{2}{9}\)+\(\frac{3}{9}\)+......\(\frac{7}{9}\)+\(\frac{8}{9}\)=?
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)
Tách 9=1+1+...+1 ( có 9 số 1)
\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)
\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)
\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )
Vậy \(A:B=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
BL:
=\(\frac{-3}{7}\).(\(\frac{2}{9}\)+\(\frac{4}{9}\)+\(\frac{8}{9}\))
=\(\frac{-3}{7}\).\(\frac{14}{9}\)
=\(\frac{-2}{3}\)
Học tôk
\(\frac{-3}{7}\cdot\frac{2}{9}+\frac{4}{9}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{8}{9}\)
\(=\frac{-3}{7}\cdot\left(\frac{2}{9}+\frac{4}{9}+\frac{8}{9}\right)\)
\(=\frac{-3}{7}\cdot\frac{14}{9}\)
\(=\frac{-2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{7}+\frac{9}{7}+\frac{5}{7}=\frac{12}{7}+\frac{5}{7}=\frac{17}{7}\)
\(\frac{8}{9}\times\frac{1}{2}=\frac{4}{9}\)
\(\frac{9}{2}:\frac{8}{6}=\frac{9}{2}\times\frac{6}{8}=\frac{27}{8}\)
Ủng hộ !
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)
\(=1.6:\frac{9^8.3-9^8.9}{9^8.\left(5+7\right)}\)
\(=6:\frac{9^8.\left(3-9\right)}{9^8.12}\)
\(=6:\frac{9^8.\left(-6\right)}{9^8.12}\)
\(=6:\left(-\frac{6}{12}\right)\)
\(=6:\left(-\frac{1}{2}\right)\)
\(=-12\)
b) 3/5 : ( -1/5-1/6)+3/5:(-1/3-16/15) ( mình chuyển về ps luôn )
=3/5: (-11/30) + 3/5 : (-7/5)
=3/5:[-11/30+(-7/5)]
=3/5:53/30
=18/53
c) (1/2-13/14):5/7-(-2/21+1/7):5/7
= -3/7:5/7-1/21:5/7
=(-3/7-1/21):5/7
=-10/21:5/7
=-2/3
câu b vá c mình làm tắt nha. chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.\frac{9}{10}\)
\(S=\frac{1}{10}\)
học tốt
S=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{9}{10}\)
S=\(\frac{1.2.3....9}{2.3.4...10}=\frac{1}{10}\)
Vậy S=\(\frac{1}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10x9/10
=9/10x(1/2+2/3)+(3/4+4/5)+(5/6+6/7)+(7/8+8/9)
=9/10x(1/3+3/5+5/7+7/9)
9/10x(1/3+3/5)+(5/7+7/9)
=9/10x1/5+5/9
9/50+5/9
=10
Bn Long làm đúng rồi bn nguyễn kim arica cứ làm theo cách đó là được .
Bn nào thấy đúng thì ủng hộ nha .
Dãy đó có số số hạng là
\(\left(\frac{8}{9}-\frac{1}{9}\right)\div\frac{1}{9}+1=\)\(8\)
Tổng dãy đó là
\(\left(\frac{1}{9}+\frac{8}{9}\right)\times8\div2=4\)
Đáp số : 4
\(\frac{1}{9}+\frac{2}{9}+\frac{3}{9}+.....+\frac{7}{9}+\frac{8}{9}\)
= \(\frac{1+2+3+4+5+6+7+8}{9}\)
= \(\frac{36}{9}=4\)