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b)
\(=\frac{3}{7}\left(\frac{9}{11}+\frac{2}{11}\right)\)
\(=\frac{3}{7}.1=\frac{3}{7}\)
`a)3/5+(-4/9)`
`=3/5-4/9`
`=27/45-20/45`
`=7/45`
`b)3/5+2/5 . 15/8`
`=3/5 + 30/40`
`=3/5+3/4`
`=12/20+15/20`
`=27/20`
`c)7/2 . 8/13 + 8/13 . (-5/2)`
`=8/13 . (7/2 +(-5)/2)`
`=8/13 . 1`
`=8/13`
`d)-5/17 . (-9/23)+9/23 . (-22/17) + 11 9/23`
`=-5/17 . (-9/23) + 9/23 . (-1) . 22/17 + 11 + 9/23`
`=-5/17 . (-9/23) + (-9/23) . 22/17+11+9/23`
`= -9/23 ( -5/17 + 22/17)+11+9/23`
`= - 9/23 . 1+11+9/23`
`=-9/23+11+9/23`
`=(-9/23+9/23)+11`
`=0+11`
`=11`
a: =27/45-20/45=7/45
b: =3/5+30/40
=3/5+3/4
=12/20+15/20
=27/20
c: =8/13(7/2-5/2+1)=8/13*2=16/13
d: =9/23*5/17-9/23*22/17+11+9/23
=-9/23+11+9/23
=11
\(-\frac{3}{5}.\frac{11}{7}+\frac{3}{-5}.\frac{4}{7}+\frac{23}{7}\)
\(=\frac{-3}{5}\left(\frac{11}{7}+\frac{4}{7}\right)+\frac{23}{7}\)
\(=\frac{-3}{5}.\frac{15}{7}+\frac{23}{7}\)
\(=\frac{-9}{7}+\frac{23}{7}=2\)
13 - 12 + 11 + 10 - 9 + 8 - 7 - 6 + 5 - 4 + 3 + 2 - 1
= 1 + 11 - 1 + 1 + 11 - 7 + 1
= 12 - 1 + 1 + 11 - 7 + 1
= 11 + 1 + 11 - 7 + 1
= 12 + 11 - 7 + 1
= 23 - 7 + 1
= 16 + 1
=17
d: =-2/7-3/11+2/7=-3/11
e: =2+3/7+1+4/7-17/7
=4-17/7=11/7
f: =-2/3*4/5+1/5*11/9
=-8/15+11/45
=-24/45+11/45=-13/45
h: =(-3,1+3,7):0,2=0,6:0,2=3
\(\dfrac{4}{5}\)+\(\dfrac{-5}{4}\)=\(\dfrac{0}{4}\)
\(\dfrac{-1}{3}\)+\(\dfrac{2}{5}-\dfrac{5}{6}\)=\(\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}\)=\(\dfrac{-23}{30}\)
\(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)=\(\dfrac{2}{3}-\dfrac{2}{5}\)=\(\dfrac{10}{15}-\dfrac{6}{15}\)=\(\dfrac{4}{15}\)
Lời giải:
$A=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+....+(297+298-299-300)+301+302-303$
$=(-4)+(-4)+(-4)+....+(-4)+300$
Số lần xuất hiện của $-4$ là:
$[(300-1):1+1]:4=75$
$A=(-4),75+300=0$
a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
(1-3)+(5-7)+(9-11)+....+(17-19)
=(-2)+(-2)+(-2)+...+(-2)
=(-2)x5
=-10