tìm n bt (1\3)^2n-1=3^5
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\(n-2⋮n+1\)
\(=>-2+n⋮1+n\)
\(=>-3+\left(1+n\right)⋮1+n\)
Do \(1+n⋮1+n\)
\(=>n+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(=>n\in\left\{-4;-2;0;2\right\}\)
\(n-2⋮n+1\)
\(n+1-3⋮n+1\)
Vì \(n+1⋮n+1\)
\(\Rightarrow n+1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
Ta cs bảng
n+1 | 1 | -1 | 3 | -3 |
n | 0 | -2 | 2 | -4 |
\(2n-3⋮n+1\)
\(2\left(n+1\right)-5⋮n+1\)
Vì \(2\left(n+1\right)⋮n+1\)
\(-5⋮n+1\)
\(\Rightarrow n+1\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}\)
Ta cs bảng
n+1 | 1 | -1 | 5 | -5 |
n | 0 | -2 | 4 | -6 |
Câu hỏi của linh tran - Toán lớp 6 - Học toán với OnlineMath
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\(n-1⋮n+2\)
\(n+2-3⋮n+2\)
\(-3⋮n+2\)
\(\Rightarrow n+2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
Tự lập bảng nha !
\(2n+7⋮n-3\)
\(2\left(n-3\right)+13⋮n-3\)
\(13⋮n-3\)
\(\Rightarrow n-3\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
Tự lập bảng nha !
\(A=\frac{15}{-\left(x+2\right)^2+5}\)
Vì \(\left(x+2\right)^2\ge0\) với mọi x
=> \(-\left(x+2\right)^2\le0\)
=>\(-\left(x+2\right)^2+5\le5\)
\(A=\frac{15}{-\left(x+2\right)^2+5}\ge\frac{15}{5}=3\)
Vậy GTNN của A là 3 khi x=-2
a) \(5^{n+3}-5^{n+1}=5^{12}.120\Leftrightarrow5^{n+1}.\left(5^2-1\right)=5^{12}.5.24\)
\(\Leftrightarrow24.5^{n+1}=5^{13}.24\Leftrightarrow5^{n+1}=5^{13}\Leftrightarrow n+1=13\Leftrightarrow n=12\)
b) \(2^{n+1}+4.2^n=3.2^7\)
\(\Leftrightarrow2^n\left(2+4\right)=3.2^7\Leftrightarrow6.2^n=3.2^7\Leftrightarrow2^n=2^6\Leftrightarrow n=6\)
c) \(3^{n+2}-3^{n+1}=486\)
\(\Leftrightarrow3^{n+1}.\left(3-1\right)=486\Leftrightarrow2.3^{n+1}=486\Leftrightarrow3^{n+1}=243\)
\(\Leftrightarrow3^n=243:3=81=3^3\Leftrightarrow n=3\)
d) \(3^{2n+3}-3^{2n+2}=2.3^{10}\)
\(\Leftrightarrow3^{2n+2}.\left(3-1\right)=2.3^{10}\)
\(\Leftrightarrow3^{2n+2}.2=2.3^{10}\Leftrightarrow3^{2n+2}=3^{10}\Leftrightarrow2n+2=10\Leftrightarrow2n=8\Leftrightarrow n=4\)
Ta có:
2n + 9 = 2n + 6 + 3
= 2(n + 3) + 3
Để (2n + 9) ⋮ (n + 3) thì 3 ⋮ (n + 3)
⇒ n + 3 ∈ Ư(3) = {-3; -1; 1; 3}
⇒ n ∈ {-6; -4; -2; 0}
(2n+9) ⋮ (n+3)
Ta có
2n + 9
= 2(n+3)3
Vì 2(n+3)3 ⋮ (n+3)
Suy ra n+3 \(\in\) Ư(3) = {-3,-1,1,3}
n+3 | -3 | -1 | 1 | 3 |
n | -6 | -4 | -2 | 0 |
Vậy n \(\in\) {0;3}
Tính các giới hạn sau:
a) lim n^3 +2n^2 -n+1
b) lim n^3 -2n^5 -3n-9
c) lim n^3 -2n/ 3n^2 +n-2
d) lim 3n -2n^4/ 5n^2 -n+12
e) lim (căn 2n^2 +3 - căn n^2 +1)
f) lim căn (4n^2-3n). -2n
(1/3)^2n-1=3^5
(1/3)^2n-1=243
(1/3)^2n.1/3=243
(1/3)^2n=243:1/3
(1/3)^2n=729
(1/3)^2n=(1/3)^-6
=>2n=-6
=>n=-6:2
n=-3
Vậy n=-3
tk mk nha!