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a)
M = 5 + 52 + 53 + ... + 560
=> 5M = 5.(5 + 52 + 53 + ... + 560)
=> 5M = 52 + 53 + 54 + ... + 561
=> 5M - M = (52 + 53 + 54 + ... + 561) - (5 + 52 + 53 + ... + 560)
=> 4M = 561 - 5
=> M = (561 - 5) : 4
a)Ta có :
\(M=5+5^2+5^3+...+5^{60}\)
\(5M=5^2+5^3+5^4+...+5^{61}\)
\(5M-M=\left(5^2+5^3+5^4+...+5^{61}\right)-\left(5+5^2+5^3+...+5^{60}\right)\)
\(4M=5^{61}-5\)
\(M=\frac{5^{61}-5}{4}\)
Ta có: 1/3 + −2/5+ 1/6 + −1/5 ≤ x < −3/4+2/7+-1/4+3/5+5/7
⇒10-12+5-6/30≤ x< -105+40-35+84+100/140
⇒-3/30≤ x <84/140
⇒-0,1≤ x < 0,6
⇒x=0
Bài 1:
b) Ta có:
\(16^5=2^{20}\)
\(\Rightarrow B=16^5+2^{15}=2^{20}+2^{15}\)
\(\Rightarrow B=2^{15}.2^5+2^{15}\)
\(\Rightarrow B=2^{15}\left(2^5+1\right)\)
\(\Rightarrow B=2^{15}.33\)
\(\Rightarrow B⋮33\) (Đpcm)
c) \(C=5+5^2+5^3+5^4+...+5^{100}\)
\(\Rightarrow C=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(\Rightarrow C=1\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{98}\left(5+5^2\right)\)
\(\Rightarrow\left(1+5^2+...+5^{98}\right)\left(5+5^2\right)\)
\(\Rightarrow C=Q.30\)
\(\Rightarrow C⋮30\) (Đpcm)
Bài 1 : a, \(A=1+3+3^2+...+3^{118}+3^{119}\)
\(A=\left(1+3+3^2+3^3\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)
\(A=\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)
\(A=1.30+...+3^{116}.30=\left(1+...+3^{116}\right).30⋮3\)
Vậy \(A⋮3\)
b, \(B=16^5+2^{15}=\left(2.8\right)^5+2^{15}\)
\(=2^5.8^5+2^{15}=2^5.\left(2^3\right)^5+2^{15}\)
\(=2^5.2^{15}+2^{15}.1=2^{15}\left(32+1\right)=2^{15}.33⋮33\)
Vậy \(B⋮33\)
c, Tương tự câu a nhưng nhóm 2 số
Bài 2 : a, \(n+2⋮n-1\) ; Mà : \(n-1⋮n-1\)
\(\Rightarrow\left(n+2\right)-\left(n-1\right)⋮n-1\)
\(\Rightarrow n+2-n+1⋮n-1\Rightarrow3⋮n-1\)
\(\Rightarrow n-1\in\left\{1;3\right\}\Rightarrow n\in\left\{2;4\right\}\)
Vậy \(n\in\left\{2;4\right\}\) thỏa mãn đề bài
b, \(2n+7⋮n+1\)
Mà : \(n+1⋮n+1\Rightarrow2\left(n+1\right)⋮n+1\Rightarrow2n+2⋮n+1\)
\(\Rightarrow\left(2n+7\right)-\left(2n+2\right)⋮n+1\)
\(\Rightarrow2n+7-2n-2⋮n+1\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in\left\{1;5\right\}\Rightarrow n\in\left\{0;4\right\}\)
Vậy \(n\in\left\{0;4\right\}\) thỏa mãn đề bài
c, tương tự phần b
d, Vì : \(4n+3⋮2n+6\)
Mà : \(2n+6⋮2n+6\Rightarrow2\left(2n+6\right)⋮2n+6\Rightarrow4n+12⋮2n+6\)
\(\Rightarrow\left(4n+12\right)-\left(4n+3\right)⋮2n+6\)
\(\Rightarrow4n+12-4n-3⋮2n+6\Rightarrow9⋮2n+6\)
\(\Rightarrow2n+6\in\left\{1;2;9\right\}\Rightarrow2n=3\Rightarrow n\in\varnothing\)
Vậy \(n\in\varnothing\)
a)
\(A=1+5+5^2+5^3+................+5^{99}\)
\(\Rightarrow5A=5+5^2+5^3+................+5^{99}+5^{100}\)
\(\Rightarrow5A-A=\left(5+5^2+5^3+.........+5^{99}+5^{100}\right)-\left(1+5+5^2+.......+5^{99}\right)\)
\(\Rightarrow4A=5^{100}-1\)
\(\Rightarrow A=\dfrac{5^{100}-1}{4}\)
Ta có :
\(A=\dfrac{5^{100}-1}{4}< B=\dfrac{5^{100}}{4}\Rightarrow A< B\)
b) Chưa có nghĩ ra!!
a, \(A=1+5+5^2+...+5^{100}\\ =>5A=5+5^2+5^3+...........+5^{101}\\ =>5A-A=\left(5+5^2+5^3+......+5^{101}\right)-\left(1+5+5^2+...5^{100}\right)\\ 4A=5^{101}-1\\ =>A=\dfrac{5^{101}-1}{4}->\left(1\right)\)
Theo đề: \(B=\dfrac{5^{101}}{4}->\left(2\right)\)
Từ (1) và (2), ta thấy: \(\dfrac{5^{101}-1}{4}< \dfrac{5^{101}}{4}\\ =>A< B\)
bài 1 mifk viết sai nha.
bài 1: cho A=1+3+3\(^2\)+3\(^3\)+...+3\(^{10}\).Tìm số tự nhiên n biết 2 x A + 1 = 3\(^n\)
B1:
\(A=1+3+3^2+3^3+...+3^{10}\\ 3A=3+3^2+3^3+3^4+...+3^{11}\\ 3A-A=3^{11}-1\\ \Rightarrow A=\frac{3^{11}-1}{2}\)
mấy câu khác tương tự nha
a) Ta có :
M= \(5+5^2+5^3+...+5^{60}\)
5M= \(5^2+5^3+5^4+...+5^{61}\)
5M - M= \(\left(5^2+5^3+5^4+...+5^{61}\right)\) - \(\left(5+5^2+5^3+...+5^{60}\right)\)
4M= \(5^{61}-5\)
M= \(\dfrac{5^{61}-5}{4}\)
thế lm sao lm đc câu c