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(-32).74-32.27+32
=32.(-74)-32.27+32
=32.[(-74)-27+1]
=32.(-100)
=-3200
\(-32\cdot74-32\cdot27+32\)
\(=32\left(-74-27+1\right)\)
\(=32\cdot\left(-100\right)\)
=-3200
\(\frac{3^4.4-3^6}{3^5.5+10.3^2.4}=\frac{3^4\left(4-3^2\right)}{3^5.5+2.5.3^2.2^2}=\frac{3^4\left(-5\right)}{3^2.5\left(3^3+2^3\right)}=\frac{3^2.3^2.5.\left(-1\right)}{3^2.5.35}=\frac{3^2\left(-1\right)}{35}=\frac{-9}{35}\)
(2^12+2^5):(2^10+2^2)
=(2^12+5):(2^10+2)
=2^17:2^12
=2^17-12
=2^5.
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\frac{12}{25}\)
\(=\frac{24}{125}\)
\(a,23\cdot125+78\cdot125-125\\ =125\cdot\left(23+78-1\right)\\ =125\cdot100\\ =12500\\ b,37\cdot46+18\cdot54-17\cdot46+54\cdot2\\ =46\cdot\left(37-17\right)+54\cdot\left(18+2\right)\\ =46\cdot20+54\cdot20\\ =20\cdot\left(46+54\right)\\ =20\cdot100\\ =2000\\ b,3\cdot2^2-2^3:2+2^2\\ =3\cdot4-4+4\\ =12\\ d,7^3\cdot9+3^2\cdot7^4\\ =7^3\cdot9+7^4\cdot9\\ =7^3\cdot9\cdot\left(1+7\right)\\ =343\cdot9\cdot8\\ =24696\)
a) \(23\cdot125+78\cdot125-125\)
\(=125\cdot\left(23+78-1\right)\)
\(=125\cdot100\)
\(=12500\)
b) \(37\cdot46+18\cdot54-17\cdot46+54\cdot2\)
\(=54\cdot\left(18+2\right)+46\cdot\left(37-17\right)\)
\(=54\cdot20+46\cdot20\)
\(=54\cdot\left(20+46\right)\)
\(=20\cdot100\)
\(=2000\)
c) \(3\cdot2^2-2^3:2+2^2\)
\(=3\cdot2^2-2^2+2^2\)
\(=3\cdot2^2\)
\(=3\cdot4\)
\(=12\)
d) \(7^3\cdot9+3^2\cdot7^4\)
\(=7^3\cdot3^2+3^2\cdot7^4\)
\(=7^3\cdot3^2\cdot\left(1+7\right)\)
\(=24696\)
Đặt \(A=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{48\cdot50}\)
\(A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
\(A=\frac{1}{10}-\frac{1}{50}=\frac{5}{50}-\frac{1}{50}=\frac{4}{50}=\frac{2}{25}\)
Vậy \(A=\frac{2}{25}\)
= \(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
= \(\frac{1}{10}-\frac{1}{50}\)= \(\frac{2}{25}\)
\(1037+\left\{743-\left[1031-\left(+57\right)\right]\right\}\)
\(=1037-1037+\left\{743-\left(+57\right)\right\}\)
\(=0+686\)
\(=686\)
\(1037+\left\{743-\left[1031-\left(+57\right)\right]\right\}\)
\(=1037+743-1031+57\)
\(=\left(743+57\right)+\left(1037-1031\right)\)
\(=800+6\)
\(=806\)
\(-32\cdot74-32\cdot27+32\)
\(=32\left(-74-27+1\right)\)
\(=32\cdot\left(-100\right)=-3200\)