Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 14:
Để số đó nhân với \(\dfrac{7}{25},\dfrac{9}{35},\dfrac{11}{40}\)là tích các số TN thì số đó phải là BCNN(25,35,40)
25=52
35=5.7
40=23.5
BCNN(25,35,40)=52.23.7=1400
Vậy số đó là 1400
Bài 16:
a) Ta có: \(A=-\dfrac{9}{5}\)
nên \(\dfrac{3x+9}{17-2x}=\dfrac{-9}{5}\)
\(\Leftrightarrow15x+45=-9\left(17-2x\right)\)
\(\Leftrightarrow15x+45+9\left(17-2x\right)=0\)
\(\Leftrightarrow15x+45+153-18x=0\)
\(\Leftrightarrow-3x=-198\)
hay x=66
b) Thay x=3 vào A, ta được:
\(A=\dfrac{3\cdot3+9}{17-2\cdot3}=\dfrac{18}{17-6}=\dfrac{18}{11}\)
\(15+4.12-3^2=15+48-9=63-9=54\\ ---\\ \left\{\left[\left(37+13\right):5\right]-45:5\right\}.7-2022^0=\left\{\left[50:5\right]-9\right\}.7-1\\ =\left\{10-9\right\}.7-1\\ =1.7-1=6\)
a) 15 + 4.12 - 3²
=15 + 48 - 9
= 63 - 9
= 54
b) {[(37 + 13) : 5].2 - 45 : 5}.7 - 2022⁰
= [(50 : 5).2 - 9].7 - 1
= (10.2 - 9).7 - 1
= (20 - 9).7 - 1
= 11.7 - 1
= 77 - 1
= 76
a, 98; 56; 24
98 = 2.72
56 = 23.7
24 = 23.3
BCNN(98; 56;24) = 23.3.72 = 1176
b, 50; 600; 120
50 = 2.52
600 = 23.3.52
120 = 23.3.5
BCNN(50; 600;120) =23.3.52= 600
A= \(\dfrac{10.11.\left(1+5.5+7.7\right)}{11.12.\left(1+5.5+7.7\right)}=\dfrac{10}{12}=\dfrac{5}{6}\)
\(A=3+3^2+...+3^{2005}\)
\(\Rightarrow3A=3^2+3^3+...+3^{2006}\)
\(\Rightarrow3A-A=3^{2006}-3\)
\(\Rightarrow2A=3^{2006}-3\)
\(\Rightarrow2A+3=3^{2006}\) là 1 lũy thừa của 3 (đpcm)
4.
\(B=1+1+2+2^2+2^3+...+2^{100}\)
\(2B=2+2+2^2+...+2^{101}\)
\(\Rightarrow2B-B=2+2^{101}-\left(1+1\right)=2^{101}\)
\(\Rightarrow B=2^{101}\) là 1 lũy thừa của 2 (đpcm)
Bài 1:
\(A=2+2^2+2^3+...+2^{2003}+2^{2004}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2002}+2^{2003}+2^{2004}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2002}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+2^4+...+2^{2002}\right)⋮7\)
Bài 2:
\(A=2+2^2+2^3+2^4+...+2^{59}+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\cdot\left(2+2^5+...+2^{57}\right)⋮15\)
Bài 3:
\(A=1+3+3^2+3^3+...+3^{1990}+3^{1991}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{1989}+3^{1990}+3^{1991}\right)\)
\(=13+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)
Bài 4:
\(A=4+4^2+4^3+4^4+...+4^{23}+4^{24}\)
\(=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{23}+4^{24}\right)\)
\(=\left(4+4^2\right)+4^2\left(4+4^2\right)+...+4^{22}\left(4+4^2\right)\)
\(=20\left(1+4^2+...+4^{22}\right)⋮20\)
a) \(\left(0,5\right)^{12}:\left(0,5\right)^{10}=\left(0,5\right)^{12-10}=\left(0,5\right)^2\)
b) \(\sqrt{36}=\pm6\)
c)\(\left(0,75\right)^{22}:\left(0,75\right)^{12}=\left(0,75\right)^{22-12}=\left(0,75\right)^{10}\)
d) \(\sqrt{49}=\pm7\)