tìm x để A là số nguyên:
A=\(\frac{x^2}{x+1}\)
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a)
\(\dfrac{13}{x-1}\in Z\\ \Rightarrow\left(x-1\right)\inƯ\left(13\right)\\ \Rightarrow\left(x-1\right)\in\left\{1;-1;13;-13\right\}\\ \Rightarrow x\in\left\{2;0;14;-12\right\}\)
b)
\(\dfrac{x+3}{x-2}=\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}\\ 1+\dfrac{5}{x-2}\in Z\\ \Rightarrow\dfrac{5}{x-2}\in Z\\ \Rightarrow\left(x-2\right)\inƯ\left(5\right)\\ \Rightarrow\left(x-2\right)\in\left\{1;-1;5;-5\right\}\\ \Rightarrow x\in\left\{3;1;7;-3\right\}\)
tham khảo
https://olm.vn/hoi-dap/detail/99049659825.html
`<=> 3 vdots x+2`
`<=> x + 2 in Ư(3)`
`@ x + 2 = 1 => x = -1`
`@ x + 2= -1 => x = -3`
`@ x + 2 = 3 => x = 1`
`@ x + 2 = -3 => x = -5`.
Vậy ...
Để A là số nguyên thì \(x+2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-1;-3;1;-5\right\}\)
a: ĐKXĐ: x<>-1
Để \(\dfrac{x^3-x^2+2}{x-1}\in Z\) thì \(x^3-x^2+2⋮x-1\)
=>\(x^2\left(x-1\right)+2⋮x-1\)
=>\(2⋮x-1\)
=>\(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;0;3;-1\right\}\)
b: ĐKXĐ: x<>2
Để \(\dfrac{x^3-2x^2+4}{x-2}\in Z\) thì \(x^3-2x^2+4⋮x-2\)
=>\(x^2\left(x-2\right)+4⋮x-2\)
=>\(4⋮x-2\)
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
c: ĐKXĐ: x<>-1/2
Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì \(2x^3+x^2+2x+2⋮2x+1\)
=>\(x^2\left(2x+1\right)+\left(2x+1\right)+1⋮2x+1\)
=>\(1⋮2x+1\)
=>\(2x+1\in\left\{1;-1\right\}\)
=>\(2x\in\left\{0;-2\right\}\)
=>\(x\in\left\{0;-1\right\}\)
\(\Leftrightarrow2-x\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\\ \Leftrightarrow x\in\left\{-9;1;3;13\right\}\)
a: Để D là số nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{4;2;8;-2\right\}\)
\(=>x^2⋮x+1=>x\left(x+1\right)-x⋮\left(x+1\right)\)
hay \(x⋮x+1\)\(=>\left(x+1\right)-1⋮x+1\)
\(=>1⋮x+1=>x=-2\)
Vậy x=-2 thì A=-4....k nha bn