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\(M=\frac{4x+5}{2x+1}=\frac{4x+2+3}{2x+1}=\frac{2\left(2x+1\right)+3}{2x+1}=\frac{2\left(2x+1\right)}{2x+1}+\frac{3}{2x+1}=2+\frac{3}{2x+1}\)
Để M là số nguyên thì \(\frac{3}{2x+1}\) là số nguyên
=>3 chia hết cho 2x+1
=>2x+1\(\inƯ\left(3\right)\)
=>2x+1\(\in\left\{-3;-1;1;3\right\}\)
=>2x\(\in\left\{-4;-2;0;2\right\}\)
=>x\(\in\left\{-2;-1;0;1\right\}\)
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a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
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Để \(x\inℤ\)khi \(\frac{3x+12}{2x+4}\)là số nguyên
hay \(3x+12⋮2x+4\Leftrightarrow6x+24⋮2x+4\)
\(\Leftrightarrow3\left(2x+4\right)+12⋮2x+4\Rightarrow2n+4\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
2n + 4 | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 6 | -6 | 12 | -12 |
n | -3/2 ( ktm ) | -5/2 ( ktm ) | -1 | -3 | -1/2 ( ktm ) | -7/2 ( ktm ) | 0 | -4 | 1 | -5 | 4 | -8 |
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a: ĐKXĐ: x<>2; x<>3
\(Q=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+1}{x-3}\)
b: Để P<1 thì P-1<0
=>\(\dfrac{x+1-x+3}{x-3}< 0\)
=>x-3<0
=>x<3
Để Q nguyên thì \(x+1⋮2x\)
\(\Leftrightarrow2x+2⋮2x\)
mà \(2x⋮2x\)
nên \(2⋮2x\)
\(\Leftrightarrow2x\inƯ\left(2\right)\)
\(\Leftrightarrow2x\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow x\in\left\{\dfrac{1}{2};-\dfrac{1}{2};1;-1\right\}\)
mà \(x\in Z\)
nên \(x\in\left\{1;-1\right\}\)
Vậy: Khi \(x\in\left\{1;-1\right\}\) thì Q nhận giá trị nguyên