(-14) mũ 5/196 mũ 2
[(-64).(-0,25) mũ 3] mũ
Ai làm nhanh mình tích
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\(\frac{2^3\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}\)
\(=\frac{2^3\cdot5^2\cdot11\cdot11\cdot7}{2^3\cdot5^2\cdot5\cdot7\cdot7\cdot11}\)
\(=\frac{11}{5\cdot7}=\frac{11}{35}\)
ta có 2^3*5^2*11^2*(7/2)^3*5^3*7^2*11
=(2^3*(7/2)^3*7^2)*(5^2*5^3)*(11^2*11)
=(2^3*7^3/2^3*7^2)*5^5*11^3
=7^5*5^5*11^3
a: Ta có: \(\frac{36^{10}\cdot14\cdot126}{35^5\cdot6}\)
\(=\frac{\left(2^2\cdot3^2\right)^{10}\cdot2\cdot7\cdot2\cdot3^2\cdot7}{\left(7\cdot5\right)^5\cdot2\cdot3}\)
\(=\frac{2^{20}\cdot3^{20}\cdot3^2\cdot2^2\cdot7^2}{7^5\cdot5^7\cdot2\cdot3}2^{21}\cdot\frac{3^{21}}{7^3}\)
b: \(\frac{21^2\cdot14\cdot126}{35^5\cdot6}\)
\(=\frac{7^2\cdot3^2\cdot2\cdot7\cdot2\cdot3^2\cdot7}{7^5\cdot5^5\cdot3\cdot2}=\frac{7^4\cdot3^4\cdot2^2}{7^5\cdot5^5\cdot3\cdot2}=\frac{2\cdot3^3}{7\cdot5^5}\)
c: \(\frac{4^9\cdot36+64^4}{100\cdot16^4}\)
\(=\frac{2^{18}\cdot2^2\cdot3^2+\left(2^6\right)^4}{2^2\cdot5^2\cdot\left(2^4\right)^4}=\frac{2^{20}\cdot3^2+2^{24}}{2^2\cdot5^2\cdot2^{16}}\)
\(=\frac{2^{20}\left(3^2+2^4\right)}{2^{18}\cdot5^2}=2^2=4\)
36 mũ 10 . 14 . 126 / 35 mũ 5 . 6 ; 21 mũ 2 . 14 .126 / 35 mũ 5 .6 ; 4 mũ 9 . 36 + 64 mũ 4 / 100 . 16 mũ 4
\(\dfrac{4^9\cdot36+64^4}{100\cdot16^4}=\dfrac{2^{20}\cdot3^2+2^{24}}{2^{18}\cdot5^2}=\dfrac{2^{20}\left(3^2+2^4\right)}{2^{18}\cdot5^2}=4\)
5.3x=405
3x=405:5
3x=81
3x=34
Vậy x=4
2x:8=4
2x=4.8
2x=32
2x=25
Vậy x=5
x28=x5
x^28-x^5=0
x^5.x^23-x^5.1=0
x^5.(x^23-1)=0
suy ra x^5=0 hoặc x^23-1=0 suy ra x^5=0^5 hoặc x^23=0+1=1 suy ra x=0 hoặc x^23=1^23 suy ra x=0 hoăc x=1
9
(x-2)^4=256
(x-2)^4=4^4
x-2=4
x=4+2=6
(x+1)^3=125
(x+1)^3=5^3
x+1=5
x=5-1=4
a) \(81^{40}=\left(3^4\right)^{40}=3^{160}\)
\(27^{14}=\left(3^3\right)^{14}=3^{42}\)
Vì \(3^{160}>3^{42}\) => \(81^{40}>27^{14}\)
b) \(5^{64}=5^{4.16}=625^{16}\)
\(3^{96}=3^{6.16}=729^{16}\)
Vì \(625^{16}< 729^{16}\)=> \(5^{64}< 3^{96}\)
c) \(125^{12}=\left(5^3\right)^{12}=5^{36}\)
\(25^{10}=\left(5^2\right)^{10}=5^{20}\)
Vì \(5^{36}>5^{20}\)=> \(125^{12}>25^{10}\)
T_i_c_k nha,mơn bạn nhìu ^^
3^x*5^x-1=224
3^x*5^x/5=224
15^x=224*5
15^x=1120
=>ko tồn tại x thỏa mãn đề bài vị 15^x luôn có tận cùng bằng 5 (x khác 0 ) hoặc 1 ( x=0) ma 1120 co tận cùng bằng 0
a) \(9^{21}.9^{33}=9^{21+33}=9^{54}\)
b) \(19^{11}.19.19=19^{11+1+1}=19^{13}\)
c) \(25^2.5^2.125=5^4.5^2.5^3=5^{4+2+3}=5^9\)
d) \(t^{2021}.t^2.\left(t^2\right)^2=t^{2021}.t^2.t^4=t^{2021+2+4}=t^{2027}\)
e) \(123^{14}:123^{13}=123^{14-13}=123\)
f) \(64^2:8^3=\left(8^2\right)^2:8^3=8^4:8^3=8^{4-3}=8=2^3\)
g) \(6^{10}:6^3:36=6^{10}:6^3:6^2=6^{10-3-2}=6^5\)
h) \(m^{20}:m^{10}.m^{10}=m^{20-10+10}=m^{20}\)
\(\frac{\left(-14\right)^5}{196^2}=\frac{-1}{196}\)
\(\left(-64\right).\left(-25\right)^3=1000000\)
thank nha