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\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)
\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)
\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)
\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)
\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)
Đến đây là tính dễ rồi :v
\((-3,2)\cdot\frac{-15}{64}+\left[0,8-2\frac{4}{15}\right]:1\frac{23}{24}\)
\(=\frac{-32}{10}\cdot\frac{-15}{64}+\left[\frac{8}{10}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-32\cdot(-15)}{10\cdot64}+\left[\frac{4}{5}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-1\cdot(-3)}{2\cdot2}+\frac{4\cdot3-34}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-22}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-517}{180}=\frac{-191}{90}\)
Bài 2 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
a) (-24) + 6 + 10 + 24
= [(-24) + 24] + 6 + 10
= 0 + 6 + 10
= 16
b) 15 + 23 + (-25) + (-23)
= [15+ (-25)] + [23 +(-23)]
= -10 + 0
= -10
c) (-3) + (-350) + (-7) + 350
=[-350 + 350] + [-3+(-7)]
= 0 + (-10)
= -10
d) (-9) + (-11) +21 + (-1)
= [ (-9) + (-11) ] + [ 21 + (-1)]
= -20 + 20
= 0
a: \(=\left(\dfrac{17}{10}+\dfrac{70}{10}-\dfrac{87}{10}\right):\left(\dfrac{23}{4}-\dfrac{11}{4}+\dfrac{9}{25}\right)\cdot\left(12,98\cdot0,25\right)+12,5\)
\(=0:\left(3+\dfrac{9}{25}\right)\cdot\left(12,98+0,25\right)+12,5\)
=12,5
b: \(=\dfrac{13}{12}\cdot\dfrac{27}{5}\cdot2\cdot\dfrac{34}{9}\cdot2\cdot\dfrac{2}{17}\)
\(=\dfrac{13}{12}\cdot2\cdot\dfrac{27}{5}\cdot\dfrac{34}{9}\cdot\dfrac{4}{17}\)
\(=\dfrac{13}{6}\cdot\dfrac{27}{5}\cdot\dfrac{8}{9}=\dfrac{8}{6}\cdot3\cdot\dfrac{13}{5}=4\cdot\dfrac{13}{5}=\dfrac{52}{5}\)
\(A=\frac{17}{23}\cdot\frac{8}{16}\cdot\frac{23}{17}\cdot\left(-80\right)\cdot\frac{3}{4}\)\(=\frac{17\cdot4\cdot2\cdot23\cdot16\cdot\left(-5\right)\cdot3}{23\cdot16\cdot17\cdot4}\)
=> \(A=\frac{2\cdot\left(-5\right)\cdot3}{1}=-30\)
\(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)
=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{7}{12}-\frac{7}{12}\right)\)
=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\cdot0=0\)
a)\(A=\frac{17}{23}.\frac{8}{16}.\frac{23}{17}.\left(-80\right).\frac{3}{4}\)
\(A=\left(\frac{17}{23}.\frac{23}{17}\right).\left(\frac{8}{16}.\frac{3}{4}\right).\left(-80\right)\)
\(A=\frac{3}{8}.\left(-80\right)\)
\(A=-30\)
b)\(C=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)
\(C=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).0\)
\(C=0\)
(-23)-(+3)x(-11)
=-23-(-33)
=10
hai dấu ngoặc đằng trước là sao