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Để P là số nguyên thì \(3x^3-5x^2+9x-15-1⋮3x-5\)
\(\Rightarrow3x-5\in\left\{1;-1\right\}\)
=>x=2(vì x là số nguyên)
Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
a) Sửa đề
\(A=\dfrac{2x^2-8}{x^2-4x+4}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)^2}=\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)^2}\)
\(A=\dfrac{2\left(x+2\right)}{x-2}\)
b) Ta có :
A = \(\dfrac{2x+4}{x-2}=\dfrac{2x-4+8}{x-2}=\dfrac{2\left(x-2\right)+8}{x-2}=\dfrac{2\left(x-2\right)}{x-2}+\dfrac{8}{x-2}\)
A = \(2+\dfrac{8}{x-2}\)
Để A có giá trị nguyên thì :
x - 2 thuộc Ư(8)
Từ đó , ta có bảng giá trị sau :
n-2 n 1 -1 2 -2 4 -4 8 -8 3 1 4 0 6 -2 10 -6
Vậy ,....
\(\dfrac{x^2-10x+25}{x^2-5x}\) <=> \(\dfrac{x^2-5x-5x+25}{x^2-5x}\)
<=> \(\dfrac{x\left(x-5\right)-5\left(x-5\right)}{x\left(x-5\right)}\)
<=> -5(x-5) >0
=> x > 5
Với các giá trị nguyên của \(x\ne-1\), để A nguyên thì \(\left(x^5+1\right)⋮\left(x^3+1\right)\)
\(\Leftrightarrow\left(x^5+x^2-\left(x^2-1\right)\right)⋮\left(x^3+1\right)\)
\(\Leftrightarrow\left(x^2\left(x^3+1\right)-\left(x^2-1\right)\right)⋮\left(x^3+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)⋮\left(x^3+1\right)\)
\(\Leftrightarrow\left(x-1\right)⋮\left(x^2-x+1\right)\)
\(\Rightarrow x\left(x-1\right)⋮\left(x^2-x+1\right)\)
\(\Leftrightarrow\left(x^2-x+1-1\right)⋮\left(x^2-x+1\right)\)
\(\Leftrightarrow1⋮\left(x^2-x+1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x+1=1\\x^2-x+1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)