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de A nguyen
=> 3 chia het cho \(\sqrt{x}-2\)
=> \(\sqrt{x}-2\in\left\{-3;-1;1;3\right\}\)
=>\(\sqrt{x}\in\left\{-1;1;3;5\right\}\)
=>x{1;1,73;1,2,23}
mình làm tròn số đấy
\(A\inℤ\Leftrightarrow3⋮\left(\sqrt{x}-2\right)\)
\(\Leftrightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Lập bảng:
| \(\sqrt{x}-2\) | \(1\) | \(-1\) | \(3\) | \(-3\) |
| \(\sqrt{x}\) | \(3\) | \(1\) | \(5\) | \(-1\) |
| \(x\) | \(9\) | \(1\) | \(25\) | Loại vì \(\sqrt{x}\ge0\) |
Vậy \(x\in\left\{1;9;25\right\}\)
\(P=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3+x\right)\left(2x+3-x\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(x-5\right)\cdot5\left(x+1\right)}-\dfrac{3\left(x+1\right)\left(x+3\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{5\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{\left(x+5\right)}{x+1}-\dfrac{x+1}{x+5}\)
\(=\dfrac{5x^2+30x+45+x^2+10x+25-x^2-2x-1}{\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{5x^2+38x+69}{\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{5x^2+38x+69}{x^2+6x+5}\)
Để P là số nguyên thì \(5x^2+30x+25+8x+34⋮x^2+6x+5\)
=>\(8x+34⋮x^2+6x+5\)
=>\(\left\{{}\begin{matrix}8x+34⋮x+1\\8x+34⋮x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x+8+26⋮x+1\\8x+40-6⋮x+5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+1\in\left\{1;-1;2;-2;13;-13;26;-26\right\}\\x+5\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\end{matrix}\right.\)
=>\(x\in\left\{-2;1\right\}\)
Lấy (1) cộng (2), ta có:
\(\left(2a+1\right)x=a^2+4a+5\)\(\Rightarrow x=\dfrac{a^2+4a+5}{2a+1}\)
Thay vào (1): \(\dfrac{\left(a^2+4a+5\right)\left(a+1\right)-10a-5}{2a+1}.\dfrac{1}{a}\)\(=\dfrac{a^3+5a^2-a}{2a+1}.\dfrac{1}{a}=\dfrac{a^2+5a-1}{2a+1}\)
Để x,y nguyên thì \(\left\{{}\begin{matrix}a^2+4a+5⋮2a+1\\a^2+5a-1⋮2a+1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a+2\right)+2a+5⋮2a+1\\a^2+2a+3a-1⋮2a+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}4⋮2a+1\\a+2⋮2a+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}4⋮2a+1\\3⋮2a+1\end{matrix}\right.\)\(\Rightarrow2a+1\in\left\{\pm1\right\}\)\(\Rightarrow a\in\left\{-1;0\right\}\)
Vậy với a=-1;0 thì hpt có nghiệm (x;y) với x,y thuộc Z.
a ) ĐK : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)\(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^{^2}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+3}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{x+4\sqrt{x}+3}\)
\(A=\frac{5}{\sqrt{x}-3}\)
A là số nguyên khi \(\sqrt{x}\)- 3 là U( 5) =-1;1;-5;5
\(\sqrt{x}\)- 3 =-1 =>x=4
\(\sqrt{x}\)-3 =1 => x =16
\(\sqrt{x}\)- 3 =-5 loại
\(\sqrt{x}\)-3 =5 => x = 64