PL
1
K
Khách
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TT
1
21 tháng 3 2017
a )
\(\frac{-1}{5}.\frac{1}{2}=\frac{\left(-1.1\right)}{5.2}=\frac{-1}{10}\)
b )
\(\frac{-1}{8}.\frac{8}{9}=\frac{-1.1}{1.9}=\frac{-1}{9}\)
c )
\(\frac{-3}{7}.\frac{14}{15}=\frac{-1.2}{1.3}=\frac{-2}{3}\)
e )
\(\left(-4\right).\frac{7}{24}=\frac{-28}{24}=\frac{-7}{6}\)
f )
\(\frac{-9}{13}.\frac{-5}{18}=\frac{\left(-1\right).\left(-5\right)}{13.2}=\frac{6}{26}=\frac{3}{13}\)
đúng đó , tk mk nhé ! hihiii
21 tháng 3 2017
a,\(\frac{-1}{10}\) b,\(\frac{-1}{9}\) c,\(\frac{-2}{5}\) e,\(\frac{-7}{6}\) f,\(\frac{-5}{26}\)
18 tháng 6 2017
Ta có : $16A=\dfrac{16}{6.10}+\dfrac{16}{7.9}+\dfrac{16}{8.8}+\dfrac{16}{9.7}+\dfrac{16}{10.6}$
$=>16A=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{6}$
$=>16A=2.(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$
$=>A=\dfrac{1}{8}(dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$
18 tháng 6 2017
\(A=\dfrac{1}{6.10}+\dfrac{1}{7.9}+\dfrac{1}{8.8}+\dfrac{1}{9.7}+\dfrac{1}{10.6}\)
\(16A=\dfrac{16}{6.10}+\dfrac{16}{7.9}+\dfrac{16}{8.8}+\dfrac{16}{9.7}+\dfrac{16}{10.6}\)
\(16A=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{6}\)
\(16A=2\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)
\(A=2:16\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)
\(A=\dfrac{1}{8}\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\left(đpcm\right)\)
7 tháng 5 2020
\(\frac{-7}{8}\cdot\frac{8}{9}=\frac{112}{x}\)
=> \(\frac{-7}{9}=\frac{112}{x}\)
=> \(\frac{-7}{9}=\frac{112\div\left(-16\right)}{x\div\left(-16\right)}\)
=> \(9=x\div\left(-16\right)\)
=> \(x=-144\)
Hoặc : \(-7x=9\cdot112\)
=> \(-7x=1008\)
=> \(x=-144\)
O
10 tháng 3 2018
\(a.-\dfrac{1}{5}.\dfrac{1}{2}\)
\(=\dfrac{-1.1}{5.2}\)
\(=\dfrac{-1}{10}.\)
\(b.-\dfrac{1}{8}.\dfrac{8}{9}\)
\(=\dfrac{-1.8}{8.9}\)
\(=\dfrac{-1}{9}.\)
\(c.-\dfrac{3}{7}.\dfrac{14}{15}\)
\(=\dfrac{-3.14}{7.15}\)
\(=\dfrac{-3.7.2}{7.3.5}\)
\(=\dfrac{-2}{5}.\)
\(d.-\dfrac{7}{5}.\dfrac{15}{21}\)
\(=\dfrac{-7.15}{5.21}\)
\(=\dfrac{-7.5.3}{5.7.3}\)
\(=-1.\)
\(e.-4.\dfrac{7}{24}\)
\(=\dfrac{-4.7}{24}\)
\(=\dfrac{-4.7}{4.6}\)
\(=\dfrac{-7}{6}.\)
\(f.-\dfrac{9}{13}.\dfrac{5}{18}\)
\(=\dfrac{-9.5}{13.18}\)
\(=\dfrac{-9.5}{13.9.2}\)
\(=\dfrac{-1.5}{13.2}\)
\(=\dfrac{-5}{26}.\)
PD
0
\(\frac{1-1+2+9+8\cdot8}{x-36.12+5}=3615\)
\(\Rightarrow\frac{1-1+2+9+64}{x-432+5}=3615\)
\(\Rightarrow\frac{75}{x-427}=3615\)
\(\Rightarrow\left(x-427\right)3615=75\)
\(\Rightarrow3615x-1543605=75\)
\(\Rightarrow3615x=1543680\)
\(\Rightarrow x=\frac{1543680}{3615}=\frac{102912}{241}\)