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a: Để A chia hết cho B thì \(\left\{{}\begin{matrix}n+1-5>0\\2-4>0\left(loại\right)\end{matrix}\right.\Leftrightarrow n\in\varnothing\)
b: \(\dfrac{A}{B}=\dfrac{5x^3y^{n+2}-3x^2y^2}{-3x^{n-1}y^n}=-\dfrac{5}{3}x^{4-n}y^2+x^{3-n}y^{2-n}\)
Để A chia hết cho B thì \(\left\{{}\begin{matrix}4-n>=0\\3-n>=0\\2-n>=0\end{matrix}\right.\Leftrightarrow n< =2\)
c: \(\dfrac{A}{B}=\dfrac{3x^6\left(2x+5\right)^{n+3}}{2x^2\left(2x+5\right)^{n-1}}=\dfrac{3}{2}x^4\left(2x+5\right)^{n+3-n+1}=\dfrac{3}{2}x^4\left(2x+5\right)^4\)
=>Với mọi N thì A chia hết cho B
a: \(A=\dfrac{4x\left(2-x\right)+8x^2}{\left(2+x\right)\left(2-x\right)}:\dfrac{x-1-2x+4}{x\left(x-2\right)}\)
\(=\dfrac{8x-4x^2+8x^2}{\left(x+2\right)\cdot\left(-1\right)\cdot\left(x-2\right)}\cdot\dfrac{x\left(x-2\right)}{-x+3}\)
\(=\dfrac{8x+4x^2}{\left(x+2\right)\cdot\left(-1\right)}\cdot\dfrac{x}{-x+3}\)
\(=\dfrac{4x\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\cdot x=\dfrac{4x^2}{x+3}\)
b: \(=\left(n^2+3n+1+1\right)\left(n^2+3n+1-1\right)\)
\(=\left(n^2+3n+2\right)\left(n^2+3n\right)\)
\(=n\left(n+1\right)\left(n+2\right)\left(n+3\right)⋮4!=24\)
Ta có: x2 – x – 12 = x2 – x – 16 + 4
= (x2 – 16) – (x – 4)
= (x – 4).(x + 4) – (x – 4)
= (x – 4).(x + 4 – 1)
= (x – 4).(x + 3)
Bài 2:
a: Để A là số nguyên thì \(3n^3+10n^2-5⋮3n+1\)
\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow n\in\left\{0;-1;1\right\}\)(do n là số nguyên)
b: Để B là số nguyên thì \(n^3-4n^2+5n-1⋮n-3\)
\(\Leftrightarrow n^3-3n^2-n^2+3n+2n-6+5⋮n-3\)
\(\Leftrightarrow n-3\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{4;2;8;-2\right\}\)
\(x^2-x+1=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(-x^2+4x-5=-\left(x^2-2.x.2+2^2\right)-1=-\left(x-2\right)^2-1< 0\forall x\)
\(a\left(2a-3\right)-2a\left(a+1\right)=a\left(2a-3-2a-2\right)=-5a⋮5\forall a\inℤ\)
a: \(\dfrac{A}{B}=\dfrac{-5}{3}x^{3-n+1}y^{n+2-n}+x^{2-n+1}y^{2-n}\)
\(=\dfrac{-5}{3}x^{2-n}y^2+x^{3-n}y^{2-n}\)
Để A chia hết cho B thì \(\left\{{}\begin{matrix}2-n\ge0\\3-n\ge0\end{matrix}\right.\Leftrightarrow n\le2\)
b: Vì n+3>n-1
nên A chia hết cho B với mọi n
Câu 1:
\(\Leftrightarrow2n^2-4n+5n-10+5⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{3;1;7;-3\right\}\)
Câu 2:
b: \(\dfrac{x^4-4x^2+2x-4a}{x-2}=\dfrac{x^4-2x^3+2x^3-4x^2+2x-4+4-4a}{x-2}\)
\(=x^3+2x^2+2+\dfrac{4-4a}{x-2}\)
Để dưlà -23 thì 4-4a=-23
=>4a=27
=>a=27/4
no trả lời