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a)13-18-(-42)-1513-18-(-42)-15
=13-18+42-1513-18+42-15
= (-18-18)+(42+42)+(-1513+13-15)
= -36+84+(-1500-15)
= -36+84+(-1515)
= 48+(-1515)
= -1467
a) chứng tỏ : abcabc chia hết cho 11
Ta có 123123:11=11193
Vậy abcabc chia hết cho 11
b)\(\frac{9\cdot15\cdot21\cdot12\cdot20}{5\cdot6\cdot45\cdot18\cdot4}=\frac{9\cdot3\cdot5\cdot3\cdot7\cdot2\cdot2\cdot3\cdot2\cdot2\cdot5}{5\cdot2\cdot3\cdot5\cdot3\cdot3\cdot2\cdot3\cdot3\cdot2\cdot2}\)\(=\frac{7\cdot5}{3}=\frac{35}{3}\)
a,-5/9 + 8/18 + -2/11 + 4/-9 + 7/15 b,5/17 + -9/15 + -2/17 + 2/5 + 14/17
= -5/9 + 4/9 + -2/+ 4/-9 + 7/15 = 5/17 + -3/5 + -2/17 + 2/5 + 14/17
=(-5/9 + 4/9 + -4/9 ) +-2/11 + 7/15 = ( 5/17 + -2/17 + 14/17) + ( 2/5 + -3/5)
= -1/3 + -2/11 + 7/15 =1 + -1/5
=-75/225 + -30/ 225 + 77/225 =4/5
= -25/225 = -1/9 ( 4/5 = 0.8 )
\(-\frac{5}{16}:0,125+1,456:\frac{7}{25}-4\frac{1}{2}\cdot0,8\)
= \(-2,5+5,2-3,6\)
= -0,9
sao cậu không tính bằng máy tính?
là \(\frac{113}{10}\)
tk nhé!
a)5×25×2×8×4×125
=(5×2)×(25×4)×(8×125)
=10×100×1000
=1000000
b)3×5^2
Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
Giải:
\(B=\dfrac{5}{18}-1,456:\dfrac{4}{15}+\dfrac{4,5.4}{5}\)
\(\Leftrightarrow B=\dfrac{5}{18}-\dfrac{273}{50}+\dfrac{18}{5}\)
\(\Leftrightarrow B=-\dfrac{356}{225}\)
Vậy ...