Tính : a, 3.7*3.8+8.3+3.7-0.4*3.7*5.
B, 2.5*9.5*4+19.5*10
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a, 2 x 4 x 8 x 0,5 x 0,125 x 0,25 x ( 0,4321 + 0,5679 )
= ( 2 x 0,5 ) x ( 4 x 0,25 ) x ( 0,125 x 8 ) x ( 0,4321 + 0,5679 )
= 1 x 1 x 1 x ( 0,4321 + 0,5679 )
= 1 x ( 0,4321 + 0,5679 )
= 1 x 1 = 1
b, 3,7 x 3,8 + 8,3 x 3,7 - 0,4 x 3,7 x 5
= 3,7 x ( 3,8 + 8,3 ) - ( 0,4 x 5 ) x 3,7
= 3,7 x 12,1 - 2 x 3,7
= 3,7 x ( 12,1 - 2 )
= 3,7 x 10,1 = 37,37
Đặt P= \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\) : \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
Có : \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)
= \(\dfrac{\left(2.5-9\right).5^{21}}{\left(5^2\right)^{10}}\)= \(\dfrac{\left(10-9\right).5^{21}}{5^{20}}\)=\(\dfrac{5^{21}}{5^{20}}\)= 5 (1)
Có: \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
= \(\dfrac{5.\left[7^{14}.\left(3.7-19\right)\right]}{\left[7^{15}.\left(3+7\right)\right]}\)=\(\dfrac{5.7^{14}.2}{7^{15}.10}\)=\(\dfrac{10.7^{14}}{7^{15}.10}\)=\(\dfrac{1}{7}\) (2)
Từ (1) và (2) suy ra:
A= 5:\(\dfrac{1}{7}\)=5.7=35
Vậy A=35 hay \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\):\(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)= 35
\(\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{5.\left[7^{14}\left(3.7-19\right)\right]}{7^{15}\left(7+3\right)}=\frac{5.7^{14}.2}{7^{15}.10}=\frac{2}{7.2}=\frac{2}{14}=\frac{1}{7}\)
a) \(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{5^{21}.\left(2.5-9\right)}{\left(5^2\right)^{10}}=\frac{5^{21}.1}{5^{20}}=5\)
b) \(\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{5.7^{14}.\left(3.7-19\right)}{7^{15}.\left(7+3\right)}=\frac{5.7^{14}.2}{7^{15}.10}=\frac{1}{7}\)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(=\frac{5^{21}.\left(2.5-9\right)}{5^{20}}:\frac{5.7^{14}\left(3.7-19\right)}{7^{15}\left(7+3\right)}\)
\(=5:\frac{5.2}{7.10}\)
\(=5:\frac{1}{7}\)
\(=35\)
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