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a)\(3.5^2-16:2^2\)
=\(3.25-16:4\)
=\(75-4=71\)
b)\(2^3.17-2^3.14\)
=\(8.17-8.14\)
\(=8\left(17-14\right)\)
\(=8.3=24\)
c)\(15.141+59.15\)
=\(15\left(141+59\right)\)
\(=15.200\)=3000
d)17.85+15.17-120
\(=17\left(85+15\right)-120\)
=\(17.100-120\)
=\(170-120=50\)
e)\(20+[30-\left(5-1\right)^2]\)
\(=20+[30-\left(4\right)^2]\)
\(=20+\left(30-16\right)\)
\(=20+14=34\)
\(a,\frac{1}{6}-\frac{2}{3}\)
\(=\frac{1}{6}-\frac{4}{6}=-\frac{3}{6}=-\frac{1}{2}\)
a: \(=36:4+2\cdot25=9+50=59\)
b: \(=79\left(82+18\right)=79\cdot100=7900\)
c: \(=49-9-\left(4^2+2^2\right)\)
\(=40-16-4=40-20=20\)
d: \(=16+\left[400:\left(200-42-138\right)\right]\)
\(=16+400:20=16+20=36\)
a) 31 . 65 + 31 . 35 - 500
=31(65+35)-500
=31..100-500
=3100-500
=2600
\(\left(-3,2\right).\frac{-15}{64}+\left(0,8-2\frac{4}{15}\right):3\frac{1}{2}=-\frac{16}{5}.-\frac{15}{64}+\left(\frac{4}{5}-\frac{34}{15}\right):\frac{7}{2}\)
\(=-\frac{3}{4}+\left(-\frac{22}{15}\right).\frac{2}{7}\)
\(=-\frac{3}{4}-\frac{44}{105}\)
\(=\frac{-315-176}{420}\)
\(=-\frac{491}{420}\)
\(\left(-3,2\right)\cdot\frac{-15}{64}+\left(0,8-2\frac{4}{15}\right):3\frac{1}{2}\)
\(=\frac{3}{4}+\frac{-22}{15}:\frac{7}{2}\)
\(=\frac{3}{4}+\frac{-44}{105}\)
\(=\frac{-315}{176}\)
=))
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2017.2019}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{2017.2019+1}{2017.2019}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2018^2}{2017.2019}\)
\(=\frac{2}{1}.\frac{2018}{2019}=\frac{4036}{2019}\)
Câu 1:
\(A=\frac{2^5.7+2^5}{2^5.3-2^5}\)= \(\frac{2^5.8}{2^5.2}\)= 4
Vậy A = 4
Câu 2:
\(B=2^3.5^3-3.\left\{400-\left[673-2^3.\left(7^8:7^6+7^0\right)\right]\right\}\)
\(B=8.125-3.\left\{400-\left[673-8.\left(7^2+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.\left(49+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.50\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-400\right]\right\}\)
\(B=1000-3.\left\{400-273\right\}\)
\(B=1000-3.127\)
\(B=1000-381\)
\(B=619\)
Vậy B = 619
\(4.\left\{3^2.\left[\left(5^2+2^3\right):11\right]-26\right\}+2009\)
\(=4.\left\{9.\left[\left(25^{ }+8\right):11\right]-26\right\}+2009\)
\(=4.\left\{9.\left[33:11\right]-26\right\}+2009\)
\(=4.\left\{9.3-26\right\}+2009\) \(=4.\left(27-26\right)+2009\)
\(=4.1+2009=2013\)
3.52-16 : 22 = 3.25 – 16 : 4 = 75 – 4 = 71