\(\frac{12}{3x7}+\frac{12}{7x11}+\frac{12}{11x15}+..........+\frac{12}{39x43}+\frac{12}{43x47}\)
Các bn làm cả cách giải giúp mh nha !!!!!!!!!!!!!!!!!!!!!!!!
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c) C=2 x5+5x8+8x11+...+23x26+26x29
d) D=3x7+7x11+11x15+...+43x47+47x51
giúp mik với thank các bn nhiều
Ta có :
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}==\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)
a, -5/12 + 1/-4 = -5/12 + -1/4 = -5/12 + -3/12 = -8/12 = -2/3
b, 5/12 + -3/28 = 35/84 + -9/84 = 26/84 = 13/42
c, -7/12 + 3/55 = -385/660 + 36/660 = -349/660
d, -5/7 + -3/4 = -20/28 + -21/28 = -41/28
Ta có \(81.\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{3}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right)\)
\(=81.\left(\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right)\)
\(=81.\left(\frac{12}{4}:\frac{5}{6}\right)\)
\(=81.\frac{18}{5}\)
\(=291,6\)
\(81\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{159}+\frac{6}{91}}\right)\)
\(=81\left(\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{2}{169}+\frac{1}{91}\right)}\right)\)
\(=81\left(3\div\frac{5}{6}\right)\)
\(=81.\frac{18}{5}\)
\(=\frac{1458}{5}\)
Đặt A = 9/4 + 9/28 +.. + 9/550
A = 9/1.4 + 9/4.7 +... + 9/22.25
A = 3( 3/1.4 + 3/4.7 + .. + 3/22.25)
A = 3 . (1/1 - 1/4 + 1/4 - 1/7 + ... +1/22 - 1/25)
A = 3 (1 - 1/25)
A = 3. 24 / 25
A = 72/25
Gọi số cần tìm là \(x\), ta có :
S = \(\frac{4}{3x7}\)+ \(\frac{4}{7x11}\)+ \(\frac{4}{11x15}\)+ ............\(x\) = \(\frac{664}{1995}\)
= \(\frac{4}{3}\)- \(\frac{4}{7}\)+ \(\frac{4}{7}\) - \(\frac{4}{11}\)+ \(\frac{4}{11}\) - \(\frac{4}{15}\)+ ..............\(x\) = \(\frac{664}{1995}\)
= \(\frac{4}{3}\)- \(x\)= \(\frac{664}{1995}\)( loại các sô giống nhau )
\(x\)= \(\frac{4}{3}\)- \(\frac{664}{1995}\)
\(x\)= \(\frac{1996}{1995}\)
a.Goi so cuoi la x ta co
....................(de bai)
=1/3-1/7+1/7-1/11+1/11-1/15+...-x=664/1995
=1/3-x=664/1995
x=1/3-664/1995
x=1/1995
lớp 4 mà học dạng này rồi sao!?
\(\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{43\cdot47}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{43\cdot47}\right)=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{47}\right)\)\(=3\left[\left(\frac{1}{3}-\frac{1}{47}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{43}-\frac{1}{43}\right)\right]=3\left[\left(\frac{47}{141}-\frac{3}{141}\right)+0+...+0\right]=3\cdot\frac{44}{141}=\frac{44}{47}\)