3n+2 + 5.3 n-1 = 864
2 n + 4.2 n+1 = 9.4 3
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2n+3 + 2n = 144
<=> 2n(23 + 1 ) = 144
<=> 2n.9 = 144
<=> 2n = 16
<=> 2n = 24
<=> n= 4
2n+ 4.2 n+1 = 9.43
<=> 2n + 2n+3 = 9.26
<=> 2n( 1 + 23 ) = 9.26
<=> 2n.9 = 9.26
<=> 2n = 26
<=> n = 6
#_W
Tìm n
a) 2n + 3 + 2n = 144
=> 2n.23 + 2n = 144
=> 2n.8 + 2n = 144
=> 2n.(8 + 1) = 144
=> 2n.9 = 144
=> 2n = 144 : 9
=> 2n = 16
=> 2n = 24
=> n = 4
Vậy n = 4
b) 2n + 4.2n + 1 = 9.43
=> 2n + 22.2n + 1 = 9.(22)3
=> 2n + 2n + 3 = 9.22.3
=> 2n + 2n.23 = 9.26
=> 2n + 2n.8 = 9.26
=> 2n.(1 + 8) = 9.26
=> 2n.9 = 9.26
=> 2n = 26
=> n = 6
Vậy n = 6
1.
\(\lim (n^3+4n^2-1)=\infty\) khi $n\to \infty$
2.
\(\lim\limits_{n\to -\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to -\infty}\frac{-\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to -\infty}\frac{-(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{-1}{3}\)
\(\lim\limits_{n\to +\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to +\infty}\frac{\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to +\infty}\frac{(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{1}{3}\)
3.
\(\lim \frac{1+2+...+n}{2n^2}=\lim \frac{n(n+1)}{4n^2}=\lim \frac{n^2+n}{4n^2}\\ =\lim (\frac{1}{4}+\frac{1}{4n})=\frac{1}{4}\)
4.
\(\lim \frac{3^n-4.2^{n-1}-10}{7.2^n+4^n}=\lim \frac{(\frac{3}{4})^n-(\frac{2}{4})^{n-1}-\frac{10}{4^n}}{7(\frac{2}{4})^n+1}\\ =\lim \frac{(\frac{3}{4})^n-(\frac{1}{2})^{n-1}-\frac{10}{4^n}}{7(\frac{1}{2})^n+1}\\ =\frac{0-0-0}{7.0+1}=0\)
a,
5n + 5n + 2 = 650
=> 5n + 5n.52 = 650
=> 5n(1 + 52) = 650
=> 5n.26 = 650
=> 5n = 25
=> n = 2
a) 5n +5n+2 = 650
5n + 5n.52 = 650
5n.(1+25 ) = 650
5n.26= 650
5n = 25 = 52
=> n = 2
b) 3n+3 +5.3n = 864
3n.33 +5.3n = 864
3n.(33+5) = 864
3n.32 = 864
3n = 27 = 33
=> n = 3
các bài cn lại bn dựa vào mak lm nha!
a) 5n + 5n+2 = 650
=> 5n+5n+2=54 +52
=> n+n+2 = 4+2
=>2n +2 = 6
=> n=2
b) 3n + 5.3n= 864
=> 3n .(1+5) = 864
=> 3n = 864 :6
=> 3n =144
=> 3n =32+33+34-3
=> n=2+3+4-3=6
c ) 5n+3 - 5n+1= (125)4 . 120
=> 5n+3 - 5n+ = 512. ( 5^3 -5)
=> n+3 -n = 12.2
=> 3=14 ( vô lí )
=> không tồn tại n
Kunzy Nguyễn: Mik ko có ý chê bạn đâu nhưng mà câu a mik thấy bạn giải có chút gọi là ''sai''!
\(\lim\dfrac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}=\lim\dfrac{\left(\dfrac{2n-1}{n}\right)\left(\dfrac{3n^2+2}{n^2}\right)^3}{\dfrac{-2n^5+4n^3-1}{n^7}}\)
\(=\lim\dfrac{\left(2-\dfrac{1}{n}\right)\left(3+\dfrac{2}{n^2}\right)^3}{-\dfrac{2}{n^2}+\dfrac{4}{n^4}-\dfrac{1}{n^7}}=-\infty\)
\(\lim3^n\left(6.\left(\dfrac{2}{3}\right)^n-5+\dfrac{7n}{3^n}\right)=+\infty.\left(-5\right)=-\infty\)
a)
\(\left(\frac{1}{3}\right)^n\cdot27^n=3^n\)
\(\Rightarrow\left(\frac{1}{3}\cdot27\right)^n=3^n\)
\(\Rightarrow9^n=3^n\)
\(\Rightarrow\left(3^2\right)^n=3^n\)
\(\Rightarrow3^{2n}=3^n\)
\(\Rightarrow2n=n\)
\(\Leftrightarrow n=0\)
Vậy \(n=0\)
d) Ta có:
\(6^{3-n}=216\)
\(\Rightarrow6^{3-n}=6^3\)
\(\Rightarrow3-n=3\)
\(\Rightarrow n=3-3\)
\(\Rightarrow n=0\)
Vậy \(n=0\)\(\text{ }\)