Tính
c) 23 41 - 15 82 . 41 25
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Giải;
1) \(347.222-222.\left(216+184\right):8\)
\(=347.222-222.400:8\)
\(=347.222-222.50\)
\(=222.\left(347-50\right)\)
\(=222.297\)
\(=65934\)
2) \(132-\left[116-\left(132-128\right).22\right]\)
\(=132-\left[116-4.22\right]\)
\(=132-\left[116-88\right]\)
\(=132-28\)
\(=104\)
3) \(16:\left\{400:\left[200-\left(37+46.3\right)\right]\right\}\)
\(=16:\left\{400:\left[200-\left(37+138\right)\right]\right\}\)
\(=16:\left\{400:\left[200-175\right]\right\}\)
\(=16:\left\{400:25\right\}\)
\(=16:16\)
\(=1\)
4) \(\left\{184:\left[96-124:31\right]-2\right\}.3651\)
\(=\left\{184:\left[96-4\right]-2\right\}.3651\)
\(=\left\{184:92-2\right\}.3651\)
\(=\left\{2-2\right\}.3651\)
\(=0.3651\)
\(=0\)
5) \(46-\left[\left(16+71.4\right):15\right]-2\)
\(=46-\left[\left(16+284\right):15\right]-2\)
\(=46-\left[300:15\right]-2\)
\(=46-20-2\)
\(=24\)
6) \(3^3.18+72.4^2-41.18\)
\(=18.\left(27-41\right)+72.16\)
\(=18.-14+1152\)
\(=-252+1152\)
\(=900\)
Giải: (tiếp)
7) \(\left(56.46-25.23\right):23\)
\(=\left(2576-575\right):23\)
\(=2001:23\)
\(=87\)
8) \(\left(28.54+56.36\right):21:2\)
\(=\left(1512+2016\right):21:2\)
\(=3528:21:2\)
\(=84\)
9) \(\left(76.34-19.64\right):\left(38.9\right)\)
\(=\left(2584-1216\right):342\)
\(=1368:342\)
\(=4\)
10) \(\left(2+4+6+...+100\right).\left(36.333-108.111\right)\)
\(=\left(2+4+6+...+100\right).\left(11988-11988\right)\)
\(=\left(2+4+6+...+100\right).0\)
\(=0\)
11) \(\left(5.4^{11}-3.16^5\right):4^{10}\)
\(=5.4^{11}:4^{10}-3.16^5:4^{10}\)
\(=5.4-3.1\)
\(=20-3\)
\(=17\)
12) \(7256.4375-725:3650+4375.7255\)
\(=4375.\left(7256+7255\right)-\dfrac{29}{146}\)
\(=4375.14511-\dfrac{29}{146}\)
\(=63485624,8\)
Câu 12 ko chắc!
\(\frac{35}{41}.\frac{-25}{21}.\frac{-82}{15}=\frac{50}{9}\)
A) \(\frac{2}{3}+\frac{1}{5}-\frac{10}{7}=\frac{13}{15}-\frac{10}{7}=\frac{-59}{105}\)
B) \(\frac{7}{12}-\frac{27}{18}.\frac{2}{18}=\frac{7}{12}-\frac{3}{2}.\frac{2}{18}=\frac{7}{12}-\frac{1}{6}=\frac{5}{12}\)
C) \(\left(\frac{23}{41}-\frac{15}{82}\right).\frac{41}{25}=\frac{31}{82}.\frac{41}{25}=\frac{31}{50}\)
D) \(\left(\frac{4}{5}+\frac{1}{2}\right).\left(\frac{3}{13}-\frac{8}{13}\right)=\frac{13}{10}.\frac{-5}{13}=\frac{-1}{2}\)
a, = \(\frac{-59}{105}\)
b = \(\frac{7}{144}\)
c= \(\frac{31}{50}\)
d=\(\frac{-1}{2}\)
\(a.\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\\ =\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{20}{21}\\ b.\dfrac{7}{12}-\dfrac{27}{7}\cdot\dfrac{1}{18}\\ =\dfrac{7}{12}-\dfrac{3}{14}=\dfrac{31}{84}\\ c.\left(\dfrac{23}{11}-\dfrac{15}{82}\right)\cdot\dfrac{41}{25}\\ =\dfrac{1721}{902}\cdot\dfrac{41}{25}=\dfrac{1721}{550}\\ d.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)\cdot\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\\ =\dfrac{13}{10}\cdot\left(\dfrac{-5}{13}\right)=\dfrac{-1}{2}\)
c) Ta có: