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\(2×\dfrac{3}{7}+\left(\dfrac{2}{9}-\dfrac{10}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)
\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-15\)
\(=\left(\dfrac{6}{7}-\dfrac{10}{7}\right)+\dfrac{2}{9}-15\)
\(=-\dfrac{4}{7}+\dfrac{2}{9}-15\)
\(=-\dfrac{22}{63}-15\)
\(=-\dfrac{967}{63}\)
\(2x.\dfrac{3}{7}+\left(\dfrac{2}{9}+\dfrac{10}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)
\(\Leftrightarrow2x\dfrac{3}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}:\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{6x}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}:\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{6x}{7}+\dfrac{2}{9}-\dfrac{10}{7}-15\)
\(\Leftrightarrow\dfrac{6x}{7}-\dfrac{1021}{63}\)
Giải:
A=5/9+2/15-6/9
=(5/9-6/9)+2/15
= -1/9 + 2/15
= 1/45
B=2/7-3/8+4/7+1/7-5/8+5/15
= (2/7+4/7+1/7) + (-3/8-5/8) +1/3
= 1+ (-1) +1/3
=1/3
C=3/5+1/15+1/57+1/3-2/9-3/4-1/36
=9/15+1/15+1/57+19/57-8/36-27/36-1/36
=(9/15+1/15)+(1/57+19/57)+(-8/36-27/36-1/36)
=2/3+20/57+(-1)
=58/57+(-1)
=1/57
D=1/1.2+1/2.3+1/3.4+...+1/99.100
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1/1-1/100
=99/100
Câu E mình ko biết làm nhé!
\(\frac{1}{7}.\frac{5}{9}+\frac{5}{9}.\frac{1}{7}+\frac{5}{9}.\frac{3}{7}\)
\(=\frac{5}{9}.\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}.\frac{5}{7}=\frac{25}{63}\)
học vui!
a) 3/7 + 4/9 + 4/7 + 5/9
= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )
= 7/7 + 9/9
= 1 + 1
= 2
b)1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
= 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
= ( 1/5 + 9/5 ) + ( 2/5 + 8/5 ) + (7/5 + 3/5 ) + ( 4/5 + 6/5 ) + 5/5
= 2 + 2 + 2 + 2 + 1
= 2 x 4 + 1
= 8 +1
= 9
c) 1/8 + 1/12 + 3/8 + 5/12
= ( 1/8 + 3/8 ) + ( 1/12 + 5/12)
= 4/8 + 6/12
= 1/2 + 1/2
= 2/4 = 1/2
mỏi tay rồi
d; (1 - \(\dfrac{1}{2}\)) x (1 - \(\dfrac{1}{3}\)) x (1 - \(\dfrac{1}{4}\)) x ... x ( 1 - \(\dfrac{1}{100}\))
= \(\dfrac{1}{2}\) x \(\dfrac{2}{3}\) x \(\dfrac{3}{4}\) x \(\dfrac{3}{4}\) x ... x \(\dfrac{99}{100}\)
= \(\dfrac{1}{100}\)
\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{7}.\dfrac{1}{9}+\dfrac{5}{9}.\dfrac{3}{7}=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{3}{7}\right)+\dfrac{5}{7}.\dfrac{1}{9}\\ =\dfrac{5}{9}.\dfrac{4}{7}+\dfrac{5}{7}.\dfrac{1}{9}=\dfrac{20}{63}+\dfrac{5}{63}=\dfrac{25}{63}\)
\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{3}{7}\)
\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{5}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}.\dfrac{9}{7}\)
\(=\dfrac{5}{7}\)
\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{1}{9}.\frac{5}{7}-\frac{5}{9}.\frac{3}{7}\)
\(N=\frac{1}{6}.\left(\frac{5}{9}+\frac{1}{9}\right).\left(\frac{2}{7}+\frac{5}{7}-\frac{3}{7}\right)\)
\(N=\frac{1}{6}.\frac{6}{9}.\frac{4}{7}\)
\(N=\frac{1}{9}.\frac{4}{7}\)
\(N=\frac{4}{63}\)
\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{1}{9}.\frac{5}{7}-\frac{5}{9}.\frac{3}{7}\)
\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{5}{9}.\frac{1}{7}-\frac{5}{9}.\frac{3}{7}\)
\(N=\frac{5}{9}.\left(\frac{1}{6}+\frac{2}{7}+\frac{1}{7}-\frac{3}{7}\right)\)
\(N=\frac{5}{9}.\frac{1}{6}\)
\(N=\frac{5}{54}\)
\(2.\frac{3}{7}\left(\frac{2}{9}.1\frac{3}{7}\right)-\frac{5}{9}:\frac{1}{9}\)
\(=2.\frac{3}{7}+\left(\frac{2}{9}-\frac{10}{7}\right)-\frac{5}{9}.9\)
\(=\frac{6}{7}+\frac{2}{9}-\frac{10}{7}+\frac{45}{9}\)
\(=\frac{-4}{7}+\frac{47}{9}\)
\(=\frac{-36}{63}+\frac{329}{63}\)
\(=\frac{293}{63}\)