5x7+5-1
Bài 1:
3/5+5/6=............... ................. ........ ......... ......... .........
6/5x7/9=................ ................. ........ ........... ........... ..
11/6-3/8=............... ............... ............ ........... .............. ........
9-3/4:5/4=.............. ................ .............. .............. .............. .............
Bài 1: Tính nhanh
5/3+5/3x5+5/5x7+...+5/101x103
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{101.103}\)
\(=\frac{2}{5}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5,7}+...+\frac{2}{101.103}\right)\)
\(=\frac{2}{5}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{2}{5}\left(1-\frac{1}{103}\right)\)
\(=\frac{2}{5}.\left(\frac{102}{103}\right)=\frac{204}{515}\)
Nhớ kiểm tra lại cho kl nhé
Tính
a. 3/(3x5) + 3/(5x7) + 3/(7x9) +... + 3/(99x101)
b. 5/(3x5) +5/(5x7) +5/(7x9) +...+ 5/(99x101)
917749738461936926399639748776398646491639394748947630373937366
1+1+1-3+0-9+8-6+5x7-5+6-9-7-5+4+7
1+1+1-3+0-9+8-6+5x7-5+6-9-7-5+4+7
=1+1+1-3+0-9+8-6+35-5+6-9-7-5+4+7
=19
bằng 19 nhé bạn ! k mình với nha !
1/5x7 + 1/7x9 + 1/9x11 + ...... + 1/2009 + 2011 + 1/X = 1/5 x 0,5
(1/1x3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13)xy=3/5
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+\frac{2}{11\times13}\right)\)
\(=\frac{1}{2}\times\left(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+\frac{11-9}{9\times11}+\frac{13-11}{11\times13}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{13}\right)=\frac{6}{13}\)
Do đó ta có:
\(\frac{6}{13}\times y=\frac{3}{5}\)
\(\Leftrightarrow y=\frac{13}{10}\).
ngu các em học quá ngu
Tính nhanh 1/(5x7) + 3/(7x13) + 5/(13x23) + 7/(23x27) + 9/(27x55)?
tìm x {x^2-[5^2-(6^2-5x7)^3-4.]^3-13.2}^3=1
a)1/1x3+1/3x5+1/5x7+...+1/Xx(x+3)=99/200
b)1/1x3+1/3x5+1/5x7+...+1/Xx(x+2)
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)
* Bạn làm theo công thức và vẫn dụng câu b nhé.