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c) 9 < 3x : 3 < 81
=> 32 < 3x - 1 < 34
=> x - 1 = {2; 3; 4}
=> x = {3; 4; 5}
d) 5x . 5x + 1 . 5 x + 2 < 218 . 518 : 218
=> 5x + x + 1 + x + 2 < 218 : 218 . 518
=> 53x + 3 < 1.518
=> 53.(x + 1) < 518
=> 3.(x + 1) < 18
=> x + 1 < 18 : 3
=> x + 1 < 6
=> x < 6 - 1
=> x < 5
c. \(9\le3^x:3\le81\)
\(\Rightarrow3^2\le3^{x-1}\le3^4\)
\(\Rightarrow3^{x-1}\in\left\{3^2;3^3;3^4\right\}\)
\(\Rightarrow x-1\in\left\{2;3;4\right\}\)
\(\Rightarrow x\in\left\{3;4;5\right\}\)
d. Thêm đk : x thuộc N
\(5^x.5^{x+1}.5^{x+2}\le2^{18}.5^{18}:2^{18}\)
\(\Rightarrow5^{x+x+1+x+2}\le5^{18}\)
\(\Rightarrow x+x+x+1+2\le18\)
\(\Rightarrow3x+3\le18\)
\(\Rightarrow3\left(x+1\right)\le18\)
\(\Rightarrow x+1\le6\)
\(\Rightarrow x\le5\)
\(\Rightarrow x\in\left\{1;2;3;4;5\right\}\)
a) 2017 + 5.[ 300 - \(\left(17-7\right)^2\)]
= 2017 + 5.[ 300 - \(10^2\)]
= 2017 + 5.[ 300 - 100]
= 2017 + 5. 200
= 2017 + 1000
= 3017
b) \(5^{27}\).5.\(5^{25}\)-|-125|
= \(5^{27}\). 5 . \(5^{25}\) - 125
= \(5^{53}\) - 125
= \(5^{53}\) - \(5^3\)
= \(5^{53}\)+ 3
c) (\(5^{25}\).18+ \(5^{15}\).7) : \(5^{17}\)
= [ (\(5^{25}\) . \(5^{15}\)) . ( 18 . 7) ] : \(5^{17}\)
= [ \(5^{40}\) . 126 ] : \(5^{17}\)
= [ \(5^{40}\) : \(5^{17}\) ] . 126
= \(5^{23}\) . 126
Phần c) chưa chắc làm đúng nha
Học tốt :'3
a) Ta có :
32006 + 32005 - 32004
= 32004 . ( 32 + 3 - 1 )
= 32004 . ( 9 + 3 -1 )
= 32004 . 11 ⋮ 11
b) Ta có ;
20061000 + 2006999
= 2006999 . ( 2006 + 1 )
= 2006999 . 2007 ⋮ 2007
a)5^36=(5^3)^12=125^12
11^24=(11^2)^12=121^12
Vi 125^12>121^12=>5^36>11^24
\(\dfrac{36^5}{18^5}=\left(\dfrac{36}{18}\right)^5\)
\(=2^5=32\)
#\(Toru\)
\(\dfrac{36^5}{18^5}\\ =\dfrac{\left(6^2\right)^5}{\left(6\cdot3\right)^5}\\ =\dfrac{6^{10}}{6^5\cdot3^5}\\ =\dfrac{6^5}{3^5}\\ =\dfrac{\left(2\cdot3\right)^5}{3^5}\\ =\dfrac{2^5\cdot3^5}{3^5}\\ =2^5=32\)