Tìm số tự nhiên n,biết rằng
a,2 mũ 2n+3=128
b,5.2 mũ n-2 mũ n=256
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Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
a) \(A=2+2^2+...+2^{2024}\)
\(2A=2^2+2^3+...+2^{2025}\)
\(2A-A=2^2+2^3+...+2^{2025}-2-2^2-...-2^{2024}\)
\(A=2^{2025}-2\)
b) \(2A+4=2n\)
\(\Rightarrow2\cdot\left(2^{2025}-2\right)+4=2n\)
\(\Rightarrow2^{2026}-4+4=2n\)
\(\Rightarrow2n=2^{2026}\)
\(\Rightarrow n=2^{2026}:2\)
\(\Rightarrow n=2^{2025}\)
c) \(A=2+2^2+2^3+...+2^{2024}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2023}+2^{2024}\right)\)
\(A=2\cdot3+2^3\cdot3+...+2^{2023}\cdot3\)
\(A=3\cdot\left(2+2^3+...+2^{2023}\right)\)
d) \(A=2+2^2+2^3+...+2^{2024}\)
\(A=2+\left(2^2+2^3+2^4\right)+\left(2^5+2^6+2^7\right)+...+\left(2^{2022}+2^{2023}+2^{2024}\right)\)
\(A=2+2^2\cdot7+2^5\cdot7+...+2^{2022}\cdot7\)
\(A=2+7\cdot\left(2^2+2^5+...+2^{2022}\right)\)
Mà: \(7\cdot\left(2^2+2^5+...+2^{2022}\right)\) ⋮ 7
⇒ A : 7 dư 2
a) \(11^n=1331\)
\(\Rightarrow11^n=11^3\)
\(\Rightarrow n=3\)
b) \(n^3=125\)
\(\Rightarrow n^3=5^3\)
\(\Rightarrow n=5\)
c) \(5^4=n\)
\(\Rightarrow625=n\)
\(\Rightarrow n=625\)
d) \(\left(n+1^2\right)=9\)
\(\Rightarrow n+1=9\)
\(\Rightarrow n=9-1\)
\(\Rightarrow n=8\)
a) 11^n = 1331
⇒ 11^n = 11^3
⇔ n = 3
b) n^ 3 = 125
⇒ n^3 = 5^3
⇔ n = 5
c) 5^4 = n
⇒ n = 625
d) ( n + 1^2 ) = 9
⇒ ( n + 1 ) = 9
⇒ n = 8
a) \(2^n:4=16\Rightarrow2^n:2^2=2^4\Rightarrow2^{n-2}=2^4\Rightarrow n-2=4\Rightarrow n=6\)
b) \(6\cdot2^n+3\cdot2^n=9\cdot2^9\)
=> \(\left(6+3\right)\cdot2^n=9\cdot2^9\)
=> \(9\cdot2^n=9\cdot2^9\Rightarrow n=9\)
c) \(3^n:3^2=243\)
=> \(3^{n-2}=3^5\)
=> n - 2 = 5 => n = 7
d) 25 < 5n < 3125
=> 52 < 5n < 55
=> n \(\in\){3;4}
em tính 3A đi
sao đok e lấy 3A-A là đc 2A
tiếp theo chéc e cx bik lm rồi nhỉ, tự lm cho quẹn
A=3+3^2+3^3+........+3^100
3A=3^2+3^3+........+3^101
3A-A=(3^2+3^3+........+3^101)-(3+3^2+3^3+........+3^100)
2A=3^101-3
suy ra: n=3^101-3+3=3^101
**** cho chị nhé! (bài này dễ, em cố gắng luyện nhìu nhé, lm hoài sẽ cok nhìu dạng nâng cao khó hơn)
Mần^o^
a)n+5 chia hết cho n-1
=>n-1+6 chia hết cho n-1
=> 6 chia hết cho n-1 hay n-1EƯ(6)={1;2;3;6}
=>nE{2;3;4;7}
b)3n+1 chia hết cho n+1
3n+3-2 chia hết cho n+1
3(n+1)-2 chia hết cho n+1
=>2 chia hết cho n+1 hay n+1EƯ(2)={1;2}
nE{0;1}
a) \(\frac{16}{2^n}=2\)
=> 2.2n = 16
=> 21+n = 24
=> 1 + n = 4
=> n = 4 - 1
=> n = 3
Vậy n = 3
b) \(\frac{\left(-3\right)^n}{81}=-27\)
=> (-3)n = -27.81
=> (-3)n = -33.34
=> (-3)n = (-3)7
=> n = 7
Vậy n = 7
c) 8n : 2n = 4
=> (8 : 2)n = 4
=> 4n = 41
=> n = 1
Vậy n = 1