38-812=38-(34)2=38-38=0
=>A=(12+23+34+45).(13+23+33+43).0=0
Vậy A=0
A chứa thừa số 38 - 812 = 38 - 38 = 0
Do đó A = 0
tính nhanh
a)\(\left(8^{2019}-8^{2018}\right):\left(8^{2016}.8^2\right)\)
b)\(\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+3^4\right).\left(3^8-81^2\right)\)
Tính :
\(\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(3^8-81^2\right)\)
\(A=-5^{22}-\left\{-222-\left[-122-\left(100-5^{22}\right)+2022\right]\right\}\)
\(B=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+3+...+20\right)\)
\(C=\dfrac{5.4^6.9^4-3^9.\left(-8\right)^4}{4.2^{13}.3^8+2.8^4.\left(-27\right)^3}\)
trình bày cách tính nhanh các phép tính sau đây
a)\(\frac{2^8\cdot6}{3^3\cdot5^4}:\frac{8^3\cdot9}{5^3\cdot3^3}-\left(2^{14}+3^{19}\right)\cdot\left(3^{81}+5^{64}\right)\left(2^4-4^2\right)\)
Thực hiện phép tính:
1, \(\left(\dfrac{-1}{2}\right)^2.\left|+8\right|-\left(-\dfrac{1}{2}\right)^3:\left|-\dfrac{1}{16}\right|\)
2, \(\left|-0,25\right|-\left(-\dfrac{3}{2}\right)^2:\dfrac{1}{4}+\dfrac{3}{4}.2017^0\)
3, \(\left|\dfrac{2}{3}-\dfrac{5}{6}\right|.\left(3,6:2\dfrac{2}{5}\right)^3\)
4, \(\left|\left(-0,5\right)^2+\dfrac{7}{2}\right|.10-\left(\dfrac{29}{30}-\dfrac{7}{15}\right):\left(-\dfrac{2017}{2018}\right)^0\)
5, \(\dfrac{8}{3}+\left(3-\dfrac{1}{2}\right)^2-\left|\dfrac{-7}{3}\right|\)
Thực hiện các phép tính (tính nhanh nhất nếu có thể):
\(A,\)\(2^2.5-\left(1^{10}+8\right):3^2\)
\(B, 5^8:5^6+4.(3^2-1)\)
\(C,\)\(400-\left\{36-20:\left[3^3-\left(8-3\right)\right]\right\}\)
a,\(3-\left(\frac{-6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
b,\(\left(-2\right)^3+2^2+\left(-1\right)^{20}+\left(-2\right)^0\)
c,\(\left(\left(3\right)^2\right)^2-\left(\left(-5\right)^2\right)^2-\left(\left(-2\right)^3\right)^2\)
d,\(2^4+8[\left(-2\right)^2:\frac{1}{2}]^0-2^{-2}.4+\left(-2\right)^2\)
e, \(2^3+3\left(\frac{1}{2}\right)^0-2^{-2}.4+[\left(-2\right)^2:\frac{1}{2}].8\)
\(\left(-2\right)^4\) .\(\left(\dfrac{5}{8}-0,25\right)\):\(\left(3\dfrac{1}{3}-2\dfrac{1}{4}\right)\)
Tối giản phân số sau bằng cách thuận tiện:
\(\frac{\left(2^4+2^2+1\right)\left(4^4+4^2+1\right)\left(6^4+6^2+1\right)\left(8^4+8^2+1\right)\left(10^4+10^2+1\right)}{\left(3^4+3^2+1\right)\left(5^4+5^2+1\right)\left(7^4+7^2+1\right)\left(9^4+9^2+1\right)\left(11^4+11^2+1\right)}\)
