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\(\frac{8^2}{7.9}.\frac{9^2}{8.10}...\frac{14^2}{13.15}\)
\(\frac{8.8}{7.9}.\frac{9.9}{8.10}...\frac{14.14}{13.15}\)
\(\frac{8.9...14}{7.8...13}.\frac{8.9...14}{9.10...15}\)
\(\frac{14}{7}.\frac{8}{15}\)
\(2.\frac{8}{15}\)
\(\frac{16}{15}\)
(8.9.10.11.12.13.14)(8.9.10.11.12.13.14)/7.8.9.10.11.12.13).(9.10.11.12.13.14.15)
=14.8/7.15
=16/15
k cho mình nhá
a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến
c)x−12−16−112−120−130−142−156=524
\(<=> x=\dfrac{5}{24}+\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(<=> x= \dfrac{13}{12}\)
\(a,\left(10\frac{2}{9}.2\frac{3}{5}\right)-6\frac{2}{9}=\frac{1196}{45}-\frac{56}{9}=\frac{1196}{45}-\frac{280}{45}=\frac{916}{45}\)
\(b,\frac{6}{7}+\frac{1}{7}.\frac{2}{7}+\frac{1}{7}.\frac{5}{7}=\frac{1}{7}\left(6+\frac{2}{7}+\frac{5}{7}\right)=\frac{1}{7}.7=1\)
\(c,3.136.8+4.14.6-14.150=3264+336-2100=1500\)
\(d,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(e,\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
a) 2/7+-3/8+11/7+1/3+1/7+5/-8
=(2/7+11/7+1/7)+(3/8+-5/8)+1/3
=2+2+1/3
=4+1/3
=13/3
b) -3/8+12/25+5/-8+2/-5+13/25
=(-3/8+-5/8)+(12/25+13/25)+-2/5
=-1+1+-2/5
=0+-2/5
=-2/5
c)7/8+1/8*3/8+1/8*5/8
=7/8+1/8*(3/8+5/8)
=7/8+1/8*1
=7/8+1/8
=1
a) 2/7+-3/8+11/7+1/3+1/7+5/-8
=(2/7+11/7+1/7)+(3/8+-5/8)+1/3
=2+2+1/3
=4+1/3
=13/3
b) -3/8+12/25+5/-8+2/-5+13/25
=(-3/8+-5/8)+(12/25+13/25)+-2/5
=-1+1+-2/5
=0+-2/5
=-2/5
c)7/8+1/8*3/8+1/8*5/8
=7/8+1/8*(3/8+5/8)
=7/8+1/8*1
=7/8+1/8
=1
a) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}\)
\(=\frac{12}{60}+\frac{-5}{60}\)
\(=\frac{7}{60}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{2}{3}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
a) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-...-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
giúp mình với