Tính bằng cách thuận tiện nhất:
c) 12,35 x 15 – 1,235 x 10 x 6 + 12,35
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a)`1,25 + 1,25 xx 2 + 7 xx 1,25=1,25 xx (1+2+7) = 1,25 xx 10 =12,5`
b)`3,62xx12–2xx0,362xx10=3,62 xx 12 -2 xx 3,62=3,62xx(12-2)=3,62xx10=36,2`
c)`12,35xx15–1,235xx10xx6+12,35=12,35xx15–12,35xx6+12,35=12,35xx(15-6+1)=12,35xx10=123,5`
d)`1,42 : 0,5 + 1,42 xx 8=1,42xx2+1,42xx8=1,42xx(2+8)=1,42xx10=14,2`
`12,35:0,25-0,35:0,25`
`=12,35xx4-0,35xx4`
`=(12,35-0,35)xx4`
`=12xx4`
`=48`
12,35 + 1,42 - 11,35 + 8,58
= (12,35 - 11,35) + (1,42 + 8,58)
= 1 + 10
= 11
12,35 + 1,42 - 11,35 + 8,58
= 12,35 + (1,42 - 11,35 ) + 8,58
= 12,35 - 9,93 + 8,58
= 2,42 + 8,58
= 11
35 x 38 + 12,35 x 4 + 12,35 x 0,5 + 12,35 x 0,125 + 51 - 12,35 x 1,5
35 x 38 + 12,35 (4+0,5+0,125+1,5)+51
= 1330+ 12,35x6,125+51
= 1330 + 75,64375+51
=1456,64375
( 6 x 8 – 48) : ( 10 + 11 + 12 + 13 + 14 + 15 )
= (48 – 48) : ( 10 + 11 + 12 + 13 + 14 + 15 )
= 0 : ( 10 + 11 + 12 + 13 + 14 + 15 )
= 0
a, \(\dfrac{5\times11\times15\times18}{6\times15\times22\times10}\)
= \(\dfrac{5\times11\times15\times6\times3}{6\times15\times11\times2\times2\times5}\)
= \(\dfrac{5\times11\times15\times6}{5\times11\times15\times6}\) \(\times\) \(\dfrac{3}{2\times2}\)
= 1 \(\times\) \(\dfrac{3}{4}\)
= \(\dfrac{3}{4}\)
b, \(\dfrac{7}{12}\) \(\times\) \(\dfrac{3}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(\dfrac{5}{12}\)
= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{7}{12}+\dfrac{5}{12}\))
= \(\dfrac{3}{5}\) \(\times\) \(\dfrac{12}{12}\)
= \(\dfrac{3}{5}\) \(\times\) 1
= \(\dfrac{3}{5}\)
\(x:100+x:10+x.9,89=12,35\\ x\times0,01+x\times0,1+x\times9,89\\ x\times\left(0,01+0,1+9,89\right)=12,35\\ x\times10=12,35\\ x=1,235\)
12,35x15-12,35x6+12,35
=12,35x(15-6+1)
=12,35x10
=123,5
`12,35 xx15-1,235xx10xx6+12,35`
`=12,35xx15-12,35xx6+12,35`
`=12,35xx(15-6+1)`
`=12,35xx10`
`=123,5`