rút gọn
5.14 mũ 15.9 mũ 9-4.3 mũ 20.9 mũ 9
_______________________________
5. 2 ,ũ 9.6 mũ 16-7. 2 mũ 29. 27 mũ 6
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(96=2\cdot48=2\cdot2\cdot24=2\cdot2\cdot2\cdot2\cdot2\cdot3=2^5\cdot3\)
\(127=127\) (do 127 là số nguyên tố)
\(260=2\cdot130=2\cdot10\cdot13=2\cdot2\cdot5\cdot13=2^2\cdot5\cdot13\)
\(312=2\cdot156=2\cdot2\cdot78=2\cdot2\cdot2\cdot39=2^3\cdot3\cdot13\)
\(860=2\cdot430=2\cdot2\cdot215=2^2\cdot5\cdot43\)
\(4\left(x-5\right)-16=24\cdot3\)
\(\Rightarrow4\left(x-5\right)-16=72\)
\(\Rightarrow4\left(x-5\right)=72+16\)
\(\Rightarrow4\left(x-5\right)=88\)
\(\Rightarrow x-5=\dfrac{88}{4}\)
\(\Rightarrow x-5=22\)
\(\Rightarrow x=22+5\)
\(\Rightarrow x=27\)
\(4.\left(x-5\right)-16=24.3\)
\(4.\left(x-5\right)-16=72\)
\(4.\left(x-5\right)=72+16\)
\(4.x-5=88\)
\(x-5=88:4\)
\(x-5=22\)
\(x=22+5\)
\(x=27\)
`#3107.101107`
A,
\(2\times32\times12+4\times6\times41+8\times27\times3\\ =24\times32+24\times41+24\times27\\ =24\times\left(32+41+27\right)\\ =24\times100\\ =2400\)
B,
\(\left(2006\times2005^{2016}-2005^{2016}\right)\div2005^{2017}\\ =\left[2005^{2016}\times\left(2006-1\right)\right]\div2005^{2017}\\ =\left(2005^{2016}\times2005\right)\div2005^{2017}\\ =2005^{2017}\div2005^{2017}\\ =1\)
`#3107.101107`
Gọi biểu thức trên là A
Ta có:
\(A=1+5^2+5^4+...+5^{40}\\ =1\cdot\left(1+5^2\right)+5^4\cdot\left(1+5^2\right)+...+5^{38}\cdot\left(1+5^2\right)\\ =\left(1+5^2\right)\cdot\left(1+5^4+...+5^{38}\right)\\ =26\cdot\left(1+5^4+...+5^{38}\right)\)
Vì \(26\cdot\left(1+5^4+...+5^{38}\right)\text{ }⋮\text{ }26\)
\(\Rightarrow A\text{ }⋮\text{ }26\)
_______
Gọi biểu thức trên là B
Ta có:
\(B=1+2^2+2^4+...+2^{100}\\ =1\cdot\left(1+2^2+2^4\right)+2^6\cdot\left(1+2^2+2^4\right)+...+2^{96}\cdot\left(1+2^2+2^4\right)\\ =\left(1+2^2+2^4\right)\cdot\left(1+2^6+...+2^{96}\right)\\ =21\cdot\left(1+2^6+...+2^{96}\right)\)
Vì \(21\cdot\left(1+2^6+...+2^{96}\right)\text{ }⋮\text{ }21\)
\(\Rightarrow B\text{ }⋮\text{ }21\)
_______
Gọi biểu thức trên là C
Ta có:
\(C=1+3^2+3^4+...+3^{100}\\ =1\cdot\left(1+3^2+3^4+3^6\right)+3^6\cdot\left(1+3^2+3^4+3^6\right)+...+3^{94}\cdot\left(1+3^2+3^4+3^6\right)\\ =\left(1+3^2+3^4+3^6\right)\cdot\left(1+3^6+...+3^{94}\right)\\ =820\cdot\left(1+3^6+...+3^{94}\right)\)
Vì \(820\cdot\left(1+3^6+...+3^{94}\right)\text{ }⋮\text{ }82\)
\(\Rightarrow C\text{ }⋮\text{ }82.\)
a) \(A=1+5^2+5^4+5^6...+5^{40}\)
\(\Rightarrow A=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{38}\left(1+5^2\right)\)
\(\Rightarrow A=26+5^4.26+...+5^{38}.26\)
\(\Rightarrow A=26\left(1+5^4+...+5^{38}\right)⋮26\)
\(\Rightarrow1+5^2+5^4+5^6...+5^{40}⋮6\left(dpcm\right)\)
b) \(B=1+2^2+2^4+2^6+...+2^{100}\)
\(\Rightarrow B=\left(1+2^2+2^4\right)+2^6\left(1+2^2+2^4\right)+...+2^{96}\left(1+2^2+2^4\right)\)
\(\Rightarrow B=21+2^6.21+...+2^{96}.21\)
\(\Rightarrow B=21\left(1+2^6+...+2^{96}\right)⋮21\)
\(\Rightarrow1+2^2+2^4+2^6+...+2^{100}⋮21\left(dpcm\right)\)
Bài C tương tự bạn tự làm nhé!
a) \(3^{a+1}=81\)
\(3^{a+1}=3^4\)
\(a+1=4\)
\(a=3\)
b) \(\left(x-1\right)^3=27\)
\(\left(x-1\right)^3=3^3\)
\(x-1=3\)
\(x=4\)
\(a,3^{a+1}=81\\ \Rightarrow3^{a+1}=3^4\\ \\ \Rightarrow a+1=4\\ \Rightarrow a=3.\\ b,\left(x-1\right)^3=27\\ \Rightarrow\left(x-1\right)^3=3^3\\ \Rightarrow x-1=3\\ \Rightarrow x=4.\)
Nếu em hỏi trên olm mà người bên hoc24 trả lời cho em thì em sẽ không thấy câu trả lời em nhá