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a: Ta có: \(2n+1⋮n+2\)
\(\Leftrightarrow2n+4-3⋮n+2\)
\(\Leftrightarrow n+2\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{-1;-3;1;-5\right\}\)
b: Để B là số nguyên thì \(n+3⋮n-2\)
\(\Leftrightarrow n-2+5⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{3;1;7;-3\right\}\)
c: Để C là số nguyên thì \(3n+7⋮n-1\)
\(\Leftrightarrow3n-3+10⋮n-1\)
\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
hay \(n\in\left\{2;0;3;-1;6;-4;11;-9\right\}\)
Bài 3:
a: Ta có: \(3x^2=75\)
\(\Leftrightarrow x^2=25\)
hay \(x\in\left\{5;-5\right\}\)
b: Ta có: \(2x^3=54\)
\(\Leftrightarrow x^3=27\)
hay x=3
Bài 2:
b: Ta có: \(30-3\cdot2^n=24\)
\(\Leftrightarrow3\cdot2^n=6\)
\(\Leftrightarrow2^n=2\)
hay n=1
c: Ta có: \(40-5\cdot2^n=20\)
\(\Leftrightarrow5\cdot2^n=20\)
\(\Leftrightarrow2^n=4\)
hay n=2
d: Ta có: \(3\cdot2^n+2^n=16\)
\(\Leftrightarrow2^n\cdot4=16\)
\(\Leftrightarrow2^n=4\)
hay n=2
a.
\(1+2+3+...+n=820\)
\(\Leftrightarrow\dfrac{n\left(n+1\right)}{2}=820\)
\(\Leftrightarrow n\left(n+1\right)=1640\)
\(\Leftrightarrow n\left(n+1\right)=40.41\)
\(\Rightarrow n=40\)
b.
\(\left(n+5\right)⋮\left(n+1\right)\)
\(\Rightarrow\left(n+1\right)+1⋮n+1\)
\(\Rightarrow n+1=Ư\left(1\right)\)
\(\Rightarrow\left[{}\begin{matrix}n+1=-1\\n+1=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}n=-2\notin N\left(loại\right)\\n=0\end{matrix}\right.\)
c.
\(\left(2n+7\right)⋮\left(n+2\right)\)
\(\Rightarrow\left(2n+4+3\right)⋮\left(n+2\right)\)
\(\Rightarrow2\left(n+2\right)+3⋮\left(n+2\right)\)
\(\Rightarrow3⋮\left(n+2\right)\)
\(\Rightarrow n+2=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
Do n tự nhiên \(\Rightarrow n\ge0\Rightarrow n+2\ge2\)
\(\Rightarrow n+2=3\)
\(\Rightarrow n=1\)