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\(A=\frac{25^3.5^5}{6.5^{10}}=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}=\frac{5^6.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
\(B=\frac{2^5.6^3}{8^2.9^2}=\frac{2^5.2^3.3^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^8.3^3}{2^6.3^4}=\frac{4}{3}\)
=2^12.3^4.(3-1)/2^12.3^5(3+1)-5^10.7^3.(1-7)/5^9.7^3.(1+2^3)
2/3.4-5.(-6)/9
=1/6-(-10/3)
1/6+10/3
7/2
= \(\frac{5^6.5^5}{6.5^{10}}\) = \(\frac{5^{11}}{6.5^{10}}\)= \(\frac{5}{6}\)
Đặt \(A=\frac{1}{4.9}+\frac{1}{9.14}++\frac{1}{14.19}+......+\frac{1}{44.49}\)
\(A=\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+.....+\frac{5}{44.49}\right)\)
\(A=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{44}-\frac{1}{49}\right)\)
\(A=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}.\frac{45}{196}=\frac{9}{196}\)
Đặt \(B=\frac{1-3-5-7-.......47-49}{89}\)
\(B=\frac{1-\left(3+5+7+......+47+49\right)}{89}\)
Từ 3 -> 49 có: (49-3):2+1=24(số hạng)
=>\(3+5+7+....+47+49=\frac{\left(49+3\right).24}{2}=624\)
=>\(B=\frac{1-624}{89}=\frac{-623}{89}=-7\)
Vậy \(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right).\frac{1-3-5-,,,,,-49}{89}=A.B=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
\(\left(\frac{3}{7}-\frac{5}{2}-\frac{3}{5}\right)-\left(-\frac{4}{7}-\frac{3}{2}+\frac{2}{5}\right)=\frac{3}{7}-\frac{5}{2}-\frac{3}{5}+\frac{4}{7}+\frac{3}{2}-\frac{2}{5}\)
\(=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{5}{2}+\frac{3}{2}\right)+\left(-\frac{3}{5}-\frac{2}{5}\right)\)
\(=1+\left(-1\right)+\left(-1\right)\)
\(=-1\)
Hok tốt nha^^
\(\frac{-3}{11}:\frac{17}{15}+\frac{-3}{11}:\frac{17}{2}+\frac{8}{11}\)
\(=\frac{-3}{11}.\frac{15}{17}+\frac{-3}{11}.\frac{2}{17}+\frac{8}{11}\)
\(=\frac{-3}{11}.\left(\frac{15}{17}+\frac{2}{17}\right)+\frac{8}{11}\)
\(=\frac{-3}{11}.1+\frac{8}{11}\)
\(=\frac{-3}{11}+\frac{8}{11}\Rightarrow\frac{5}{11}\)
K mik nhe
\(\frac{-3}{11}:\frac{17}{15}+\frac{-3}{11}:\frac{17}{2}+\frac{8}{11}\)
\(=\frac{-3}{11}.\left(\frac{15}{17}+\frac{2}{17}\right)+\frac{8}{11}\)
\(=\frac{-3}{11}.1+\frac{8}{11}\)
=\(\frac{-3}{11}+\frac{8}{11}\)
\(=\frac{5}{11}\)
\(\frac{2^5.6^3}{8^2.9^2}\) = \(\frac{2^5.2^3.3^3}{2^6.3^4}\) = \(\frac{2^8.3^3}{2^6.3^4}\) = \(\frac{2^2}{3}\) = \(\frac{4}{3}\)
bâm máy tính đi