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a) 42.50 – 38.50 – 25.4 = 50.(42 – 38) – 25.4 = 50.4 – 25.4 = (50 – 25).4 = 25.4 = 100
b) 32.38 + 17.94 – 15.38 – 32.17 = 38.(32 – 15) + (94 – 32).17 = 38.17 + 62.17 = 17.(38+62) = 17.100 = 1700
\(a,=27\cdot20+73\cdot20-27=20\left(27+73\right)-27\\ =20\cdot100-27=2000-27=1973\\ b,=42\cdot50-38\cdot50-50\cdot2=50\left(42-38-2\right)=50\cdot2=100\\ e,=254\left(58+21\cdot2\right)=254\cdot100=25400\\ c,=53\left(12+172-84\right)=53\cdot100=5300\\ d,=24\left(31+42+27\right)=24\cdot100=2400\\ f,=\left(9-8-...-1\right)\left(500\cdot9-500\cdot9\right)=\left(9-8-...-1\right)\cdot0=0\)
\(D=12\cdot\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{22}+...+\frac{1}{97}-\frac{1}{202}\right)\)
\(D=12\cdot\left(\frac{1}{6}-\frac{1}{202}\right)\)
\(D=12\cdot\frac{49}{303}\)
\(D=\frac{588}{303}\)
a, 37.12=37.3.4=111.4=444
b, 15873.21= 15873. 7. 3= 111111. 3= 333333
a, Ta có: 37 . 3 = 111
mà \(12=3.4\)
\(\Rightarrow37.12=37.3.4=111.4=444\)
b, Ta có: 15873.7 = 111111
mà \(21=7.3\)
\(\Rightarrow15873.21=15873.7.3=111111.3=333333\)
Bài 3 :
A = 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33
=> A = ( 33 + 26 ) . 8 : 2 = 236
Vậy A = 236
\(\text{#Hok tốt!}\)
a) 2 . 31 . 12 + 4 . 6 . 42 + 8 . 27 . 3
= 24 . 31 + 24 . 42 + 24 . 27
= 24 . ( 31 + 42 + 27 )
= 24 . 100
= 2400
tính nhanh
357,8 x ( 9+20-19)=3578
15 x 9+15+9-1=15 x 10 +8 =158
a ) 21 + 369 + 79 = ( 21 + 79 ) + 369
= 100 + 369
= 469
b ) 154 + 87 + 246 = ( 154 + 246 ) + 87
= 400 +87
= 487
1/a) 12 - x= 1-(-5)
12 - x = 6
x= 12-6
x=6
b)| x+4|= 12
x+4 = \(\pm\)12
*x+4=12
x=8
*x+4= -12
x=-16
2/Tìm n
\(n-5⋮n+2\)
=> \(n+2-7⋮n+2\)
mà \(n+2⋮n+2\)
=> 7\(⋮\)n+2
=> n+2 \(\varepsilon\)Ư(7)= {1;-1;7;-7}
n+2 | 1 | -1 | 7 | -7 |
n | -1 | -3 | 5 | -9 |
3/a)4.(-5)2 + 2.(-12)
= 2.2.(-5)2 + 2.(-12)
=2[2.25.(-12)]
=2.(-600)
=-1200
\(\dfrac{12}{6.7}+\dfrac{12}{7.22}+\dfrac{12}{22.15}+\dfrac{12}{15.38}+...+\dfrac{12}{97.202}\)
\(=12.\left(\dfrac{1}{6.7}+\dfrac{1}{7.22}+\dfrac{1}{22.15}+\dfrac{1}{15.38}...+\dfrac{1}{97.202}\right)\)
\(=12.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{15}+...+\dfrac{1}{97}-\dfrac{1}{202}\right)\)
\(=12.\left(\dfrac{1}{6}-\dfrac{1}{202}\right)\)
\(=\dfrac{196}{101}\)
ko phải