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a) Ta có: n+4 chia hết cho 4.
Suy ra 4 chia hết cho n.Vậy n=1;2
b, 3n+7 chia hết cho n => 7 chia hết n
Vậy n=1
còn nhiều quá
a)\(\left(2n+1\right)^3=27\Rightarrow\left(2n+1\right)^3=3^3\)
\(\Rightarrow2n+1=3\)
\(n=\frac{3-1}{2}=\frac{2}{2}=1\)
Vậy \(n=1\)
b)\(\left(n-2\right)^2=n-n\Leftrightarrow\left(n-2\right)^2=0\)
\(\left(n-2\right)^2=0^2\Leftrightarrow n-2=0\)
\(\Leftrightarrow n=0+2=2\)
Vậy \(n=2\)
a) (2n + 1)^3 = 27
=>(2n + 1)^3 = 3^3
=>2n+1=3
=> 2n=2
=>n=1
b) (n-2)^2 = n-n
=>(n-2)^2=0
=>n-2=0
=>n=2
a)
\(\left(2n+1\right)^3=27\)
\(\left(2n+1\right)^3=3^3\)
\(2n+1=3\)
\(2n=3+1\)
\(2n=4\)
\(n=4\div2\)
\(n=2\)
b)
\(\left(n+2\right)^2=\left(n+2\right)^4\)
\(\left(n+2\right)^4-\left(n+2\right)^2=0\)
\(\left(n+2\right)^2\cdot\left(n+2\right)^2-\left(n+2\right)^2\cdot1=0\)
\(\left(n+2\right)^2\cdot\left[\left(n+2\right)^2-1\right]=0\)
\(\Rightarrow\left(n+2\right)^2=0hoạc\left(n+2\right)^2-1=0\)
\(\left(n+2\right)^2=0\)
\(n+2=0\)
\(n=0+2\)
\(n=2\)
\(\left(n+2\right)^2-1=0\)
\(\left(n+2\right)^2=0+1\)
\(\left(n+2\right)^2=1\)
\(n+2=1\)
\(n=1+2\)
\(n=3\)
Vậy \(n\in\left\{2;3\right\}\)
a) (2n + 1)3 = 33
\(\Rightarrow\)2n + 1 = 3
2n = 3 - 1
2n = 2
n = 2 : 2
n = 1
a) (2n+1)^3=3^3
(2n+1)=3
2n=2
n=1
b) (n-2)^2=(n-2)^4
(n-2)^2=1
n-2 =1 hoặc n-2 =-1
n=3 hoặc n =1
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, 2 + 4 + 6 + … + 2n = 2 + 2 n n 2 = n(n+1)
Ta có n(n+1) = 210. Ta phân tích số 210 ra thừa số nguyên tố rồi ghép các thừa số lại để được tích của hai số tự nhiên liên tiếp.
210 = 2.3.5.7 = (2.7).(3.5) = 14.15
n(n+1) = 14.15
Vậy n = 14
b, 1 + 3 + 5 +…+ (2n – 1) = 1 + 2 n - 1 2 = n 2
Ta có: n 2 = 225 n 2 = 3 2 . 5 2 = 15 2
=> n = 15
Vậy n = 15
a) 2 + 4 + 6 + ... + 2n = 210
1.2 + 2.2 + 2.3 + ... + 2n = 210
2.(1+2+3+...+n) = 210
1 + 2 + 3 + ... + n = 105
\(\frac{n\left(n+1\right)}{2}\)= 105
n(n+1) = 210
n(n+1) = 14.15
=> n = 14
b) 1+3+5+...+(2n-1)=225
\(\frac{\left(2n-1+1\right).n}{2}\) =225
\(\frac{2n.n}{2}\) =225
\(\frac{2.n^2}{2}\) =225
\(n^2\) =225
Ta có: \(n^2\) =225 = \(3^2\).\(5^2\)= \(\left(15\right)^2\)
=> n = 15
