1/3 . 1/7 + 1/7 . 1/11 + 1/11 . 1/15 + ...+ 1/2011 . 1/2015
đề là rút gọn C .
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\(C=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{13}+...+\frac{1}{2011}.\frac{1}{2015}\)
\(C=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.13}+...+\frac{1}{2011.2015}\)
\(4C=4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.13}+...+\frac{1}{2011.2015}\right)\)
\(4C=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.13}+...+\frac{4}{2011.2015}\)
\(4C=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{2011}-\frac{1}{2015}\)
\(4C=\frac{1}{3}-\frac{1}{2015}=\frac{2012}{6045}\)
\(C=\frac{2012}{6045}:4=\frac{503}{6045}\)
\(C=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+....+\dfrac{1}{2011}-\dfrac{1}{2015}\right)\)
\(C=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{2015}\right)=\dfrac{1}{4}\left(\dfrac{2012}{3.2015}\right)=\dfrac{503}{3.2015}\)
b1 -10/14
b2 -4/5
b3
a 2/9-7/8.x=1/3
7/8.x=2/9-1/3=-1/9
x=-1/9:-7/8=8/63
b 23/7.x-1/8=11/4
23/7.x=11/4+1/8=23/8
x=23/8:23/7=7/8
b4
Quyển truyện cs số trang:
36:(1−1/4−9/20)=120(trang)
\(B=\frac{0,6-\frac{3}{11}+\frac{3}{13}}{1,4-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(B=\frac{\frac{3}{5}-\frac{3}{11}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(B=\frac{3\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(B=\frac{3}{5}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(B=\frac{3}{5}-\frac{2}{7}=\frac{11}{35}\)
a) \(\frac{1.3+3.5+5.7+7.9}{3.6+9.10+15.14+21.18}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.2.3+3.5.2.3+5.7.2.3+7.9.2.3}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.6+3.5.6+5.7.6+7.9.6}\)
= \(\frac{1.3+3.5+5.7+7.9}{6.\left(1.3+3.5+5.7+7.9\right)}=\frac{1}{6}\)
Dấu "." là dấu nhân cấp 2
b) \(\frac{1.2+2.3+3.4+4.5}{3.6+6.9+9.12+12.15}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.3.3+2.3.3.3+3.4.3.3+4.5.3.3}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.9+2.3.9+3.4.9+4.5.9}\)
= \(\frac{1.2+2.3+3.4+4.5}{9.\left(1.2+2.3+3.4+4.5\right)}=\frac{1}{9}\)
Dấu "." là dấu nhân cấp 2
c) \(\frac{0,3+\frac{3}{7}+\frac{3}{11}}{0,4+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{\frac{3}{10}+\frac{3}{7}+\frac{3}{11}}{\frac{4}{10}+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{3.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}{4.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}=\frac{3}{4}\)
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
bài 5
a)
`7/5xx15/14`
`=105/70=3/2`
b)
`8/15xx25/8`
`=200/120=5/3`
c)
`2/15xx5/22`
`=10/330`
`=1/33`
bài 6
chiều rộng cx là đơn vị m phải ko ạ?
diện tích căn phòng là
`11/2xx7/2=77/4(m^2)`