Tính bằng cách hợp lý:
a. (-125).4.25.11.(-8)
b. (-123).(2-120)-120.123
c. 8.(-3)+8.(-7)
d. (-12).(-3)-7.(-12)
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b: \(=128\left(-25-89+89-125\right)=128\cdot\left(-150\right)=-19200\)
a) 5 . 125 . 2 . 41 . 8
= (5 . 8) . (125 . 2) . 41
= 40 . 250 . 41
= 10 000 . 41
= 410 000
b) 25 . 7 . 10 . 4
= (25 . 4 ) . (10 . 7)
= 100 . 70
= 7000
Chúc bạn học tốt!! ^^
a,5.125.2.41.8=(125.8).(2.5).41=1000.10.41=410000
b,25.7.10.4=(25.4).70=100.70=7000
c,8.12.125.2=(8.125).12.2=100.24=2400
d,4.36.25.50=(4.25).(2.50).18=100.100.18=180000
a)= \(2^3\left(17-14\right)\)
\(=8.3\)
\(=24\)
b)\(=17\left(85+15\right)-120\)
\(=17.100-120\)
\(=1700-120\)
\(=1580\)
c)\(=\left(25.4\right).\left(125.8\right).\left(2.5\right)\)
\(=100.1000.10\)
\(=1000000\)
d)\(=24.53+24.87-24.40\)
\(=24\left(53+87-40\right)\)
\(=24.100\)
\(=2400\)
e)\(=5.7.77-7.60-7.7.25-15.6.7\)
\(=7\left(5.77-60-7.25-15.6\right)\)
\(=7\left(385-60-175-90\right)\)
\(=7.60\)
\(=420\)
a, \(7^{2019}=7^3.7^{2016}=343.\left(7^4\right)^{504}=343.\left(...1\right)^{504}=343.\left(...1\right)=\left(....3\right)\)
b, \(12^{2019}=12^3.12^{2016}=\left(...8\right).\left(12^4\right)^{504}=\left(...8\right).\left(...6\right)^{504}=\left(...8\right).\left(...6\right)=\left(...8\right)\)
c, \(11^8+12^8+13^8+14^8+15^8+16^8\)
\(=\left(...1\right)+\left(12^4\right)^2+\left(13^4\right)^2+\left(...5\right)+\left(...6\right)=\left(....1\right)+\left(....6\right)^2+\left(...1\right)^2+\left(...5\right)+\left(...6\right)\)
\(=\left(....1\right)+\left(....6\right)+\left(...1\right)+\left(...5\right)+\left(...6\right)=\left(...9\right)\)
d, \(11^{123}+13^{124}+15^{125}=\left(....1\right)+\left(13^4\right)^{31}+\left(....5\right)=\left(...1\right)+\left(...1\right)+\left(....5\right)=\left(...7\right)\)
\(a,7^{2019}=\left[7^3\right]^{673}=\overline{....}3^{673}=\overline{....3}\)
Vậy chữ số tận cùng của 72019 là 3
\(b,12^{2019}=\left[12^3\right]^{673}=\overline{....8}^{673}=\overline{....8}\)
Vậy chữ số tận cùng của 122019 là 8
Làm nốt
a) (-125) . 4
= -500
b) (-2017) . (-2016) . 0
= 0
c) (-12) . (-11)
= 132
d) (-4) . (-8) . (-25) . (-125)
= [(-4).(-25)].[(-8).(-125)]
= 100.1000
= 100000
e) (-2).(-5).(-7)
= [(-2).(-5)].(-7)
= 10.(-7)
= -70
a: \(\left(-125\right)\cdot4=-\left(125\cdot4\right)=-500\)
b: \(\left(-2017\right)\cdot\left(-2016\right)\cdot0=0\cdot2016\cdot2017=0\)
c: \(\left(-12\right)\cdot\left(-11\right)=12\cdot11=12\cdot10+12=120+12=132\)
d: \(\left(-4\right)\cdot\left(-8\right)\cdot\left(-25\right)\cdot\left(-125\right)\)
\(=4\cdot8\cdot25\cdot125\)
\(=4\cdot25\cdot8\cdot125\)
\(=100\cdot1000=100000\)
e: \(\left(-2\right)\cdot\left(-5\right)\cdot\left(-7\right)\)
\(=-7\cdot2\cdot5\)
\(=-7\cdot10=-70\)
A = (2 + 22) + (23 + 24 ) +…(2199 + 2200)
A = 6 + 22 (2 + 22 ) +… + 2198 (2 + 22)
A = 6 + 22 (6 ) +… + 2198 (6)
A = 6(1 + 22 +… + 2198)
Vậy A chia hết cho 6
a) \(\left(\frac{2}{7}+\frac{3}{8}+\frac{12}{7}+\frac{5}{8}\right).\frac{19}{36}\)
\(=\left\{\left(\frac{2}{7}+\frac{12}{7}\right)+\left(\frac{3}{8}+\frac{5}{8}\right)\right\}.\frac{19}{36}\)
\(=\left(\frac{14}{7}+\frac{8}{8}\right).\frac{19}{36}\)
\(=\left(2+1\right).\frac{19}{36}\)
\(3.\frac{19}{36}=\frac{19}{12}\)