C=\(\dfrac{10}{7x12}\)+\(\dfrac{10}{12x17}\)+....\(\dfrac{10}{502x507}\)
a)\(\dfrac{3}{5x8}+\dfrac{3}{8x11}+...+\dfrac{3}{2006x2009}\)
b)\(\dfrac{1}{6x10}+\dfrac{1}{10x14}+...+\dfrac{1}{402x406}\)
c)\(\dfrac{10}{7x12}+\dfrac{10}{12x17}+...+\dfrac{10}{502x507}\)
Chú ý: x ở đề bài là dấu nhân nha các bạn.Mong các bạn giúp mk nhanh 1 chút vì mk đag cần gấp
a)
\(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2006.2009}\)
\(=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+....+\frac{2009-2006}{2006.2009}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)
b)
\(B=\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{402.406}\)
\(\Rightarrow 4B=\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\)
\(4B=\frac{10-6}{6.10}+\frac{14-10}{10.14}+...+\frac{406-402}{402.406}\)
\(4B=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{402}-\frac{1}{406}\)
\(4B=\frac{1}{6}-\frac{1}{406}=\frac{100}{609}\Rightarrow B=\frac{25}{609}\)
c)
\(C=\frac{10}{7,12}+\frac{10}{12.17}+...+\frac{10}{502.507}\)
\(\Rightarrow \frac{C}{2}=\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{502.507}\)
\(\frac{C}{2}=\frac{12-7}{7.12}+\frac{17-12}{12.17}+...+\frac{507-502}{502.507}\)
\(\frac{C}{2}=\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{502}-\frac{1}{507}\)
\(\frac{C}{2}=\frac{1}{7}-\frac{1}{507}=\frac{500}{3549}\)
\(\Rightarrow C=\frac{1000}{3549}\)
a) A=10/7x12+10/12x17+10/17x22+.....+10/502x507
\(A=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+...+\frac{10}{502.507}\)
\(=2\left(\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+...+\frac{5}{502.507}\right)\)
\(=2\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+...+\frac{1}{502}-\frac{1}{507}\right)\)
\(=2\left(\frac{1}{7}-\frac{1}{507}\right)\)
\(=tự tính\)
a)\(\frac{3}{5x8}+\frac{3}{8x11}+...+\frac{3}{2006x2009}\)
b)\(\frac{1}{6x10}+\frac{1}{10x14}+...+\frac{1}{402x406}\)
c)\(\frac{10}{7x12}+\frac{10}{12x17}+...+\frac{10}{502x507}\)
Ai làm nhanh nhất mk sẽ tick cho.Mk đag cần gấp.Mong các bn giúp đỡ
a,A=1/5-1/8+1/8-1/11+...+1/2006-1/2009=1/5-1/2009=2004/10045
b,B=1/4x(4/6x10+4/10x14+...+4/402x406)
=1/4x(1/6-1/10+1/10-1/14+...+1/402-1/406)
=1/4x(1/6-1/406)
=1/4x100/609=25/609
c,C=2x(5/7x12+5/12x17+...+5/502x507)
=2x(1/7-1/12+1/12-1/17+...+1/502-1/507)
=2x(1/7-1/507)
=2x500/3549
=1000/3549
Xin lỗi vì ko viết được rõ ràng.Mong bạn thông cảm. Chúc bạn học tốt.
\(\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{2006\times2009}\)
\(=\frac{1}{3}\left(\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{2006\times2009}\right)\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2006}-\frac{1}{2009}\right)\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{2009}\right)\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{2009}\right)\)
\(=\frac{1}{3}\left(\frac{2009}{10045}-\frac{5}{10045}\right)\)
\(=\frac{1}{3}.\frac{2004}{10045}=\frac{2004}{30135}\)
C= \(\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)
\(=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)
\(=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)
\(=\dfrac{5}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)
\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{5}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{14}\)
Cho đẳng thức: (-8).15=12.(-10) Tỉ lệ thức đc suy ra từ đẳng là:
A.\(\dfrac{-18}{12}\)=\(\dfrac{15}{-10}\)
B.\(\dfrac{15}{12}\)=\(\dfrac{-10}{-8}\)
C.\(\dfrac{-8}{-10}\)=\(\dfrac{15}{12}\)
D.\(\dfrac{-8}{15}\)=\(\dfrac{12}{-10}\)
Mong mn giúp đỡ.
Mk đang cần gấp.
Thanks:)
\(\text{B.}\dfrac{15}{12}=\dfrac{-10}{-8}\)
Không quy đồng ,hãy so sánh hai phân số
a \(\dfrac{19}{10}và\dfrac{10}{11}\)
b \(\dfrac{11}{10}và\dfrac{12}{11}\)
c \(\dfrac{9}{10}và\dfrac{10}{11}\)
a. 19/10 > 10/11
b. 11/10 = 12/11
c. 9/10 = 10/11
a)\(\dfrac{19}{10}>\dfrac{10}{11}\)
b)\(\dfrac{11}{10}=\dfrac{12}{11}\)
c)\(\dfrac{9}{10}< \dfrac{10}{11}\)
Tính giá trị các biểu thức:
A=(-1)+(-5)+(-9)+...+(-101)
B=\(\dfrac{-5}{17}\).\(\dfrac{8}{19}\)+\(\dfrac{-12}{17}\). \(\dfrac{8}{19}\) - \(\dfrac{11}{19}\)
C=\(\dfrac{10}{1.6}\)+\(\dfrac{10}{6.11}\)+\(\dfrac{10}{11.16}\)+...+\(\dfrac{10}{2016.2021}\)
`#lv`
`A=(-1)+(-5)+(-9)+...+(-101)`
`=-(1+5+9+...+101)`
Số số hạng là :
`[101-(-1)]:4+1=26(` số hạng `)`
Tổng là :
`[(-101)+(-1)]xx26:2=-1326`
Vậy `A=-1326`
__
`B=-5/17 . 8/19 + (-12)/17 . 8/19 - 11/19`
`=((-5)/17+(-12)/17).8/19-11/19`
`=-1.8/19-11/19`
`=-8/19-11/19`
`=-8/19+(-11)/19`
`=-19/19`
`=-1`
__
`C=10/1.6 + 10/6.11 + 10/11.16 + ... + 10/2016.2021`
`=2.(1-1/6+1/6-1/11+...+1/2016-1/2021)`
`=2(1-1/2021)`
`=2. (2021/2021-1/2021)`
`=2. 2020/2021`
`=4040/2021`
Tính rồi rút gọn (theo mẫu):
Mẫu: \(\dfrac{9}{10}-\dfrac{4}{10}=\dfrac{9-4}{10}=\dfrac{5}{10}=\dfrac{1}{2}\) |
a) \(\dfrac{15}{8}-\dfrac{13}{8}\) b) \(\dfrac{7}{15}-\dfrac{2}{15}\) c) \(\dfrac{11}{12}-\dfrac{2}{12}\) d) \(\dfrac{19}{7}-\dfrac{5}{7}\)
a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
So sánh A; B; C biết
A = \(\left(-\dfrac{43}{51}\right).\left(-\dfrac{19}{80}\right)\)
B = \(\left(-\dfrac{7}{13}\right).\left(-\dfrac{4}{65}\right).\left(-\dfrac{8}{31}\right)\)
C = \(\dfrac{-5}{10}.\dfrac{-4}{10}.\dfrac{-3}{10}...\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\)
\(A=\left(-\dfrac{43}{51}\right)\left(-\dfrac{19}{80}\right)\)
=>A>0(1)
\(B=\left(-\dfrac{7}{13}\right)\left(-\dfrac{4}{65}\right)\left(-\dfrac{8}{21}\right)\)
=>B<0(2)
C\(=-\dfrac{5}{10}.\left(-\dfrac{4}{10}\right).....\left(\dfrac{4}{10}\right)\left(\dfrac{5}{10}\right)=0\)
=>C=0(3)
Từ 1;2;3 =>A>C>B
\(A=\dfrac{-43}{51}.\dfrac{-19}{80}\Leftrightarrow A>0\left(1\right)\)
\(B=\left(\dfrac{-7}{13}\right).\left(-\dfrac{4}{65}\right).\left(\dfrac{-8}{31}\right)\Leftrightarrow B< 0\left(2\right)\)
\(C=\dfrac{-5}{10}.\dfrac{-4}{10}...........\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\Leftrightarrow C=0\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow A>C>B\)