Question

[Nguyễn Thị Hạnh]

Bài 1. (2,5 điểm)

1. Giải hệ phương trình:  $\left\{ \begin{aligned}  & 2x+3y=-2 \\  & 4x+y=1 \end{aligned} \right.$

2. Cho biểu thức: $P=\Big( \dfrac{x}{x\sqrt{x} - 4\sqrt{x}} - \dfrac{6}{3\sqrt{x} -6} + \dfrac{1}{\sqrt{x}+2} \Big) : \Big(\sqrt{x} - 2 + \dfrac{10-x}{\sqrt{x}+2} \Big)$ với $x>0;\, x \ne 0$.

a) Rút gọn biểu thức $P$.

b) Tìm các giá trị nguyên của $x$ để giá trị của biểu thức $Q = (-\sqrt{x}-1).P$ đạt giá trị nguyên.

Answer 1

[Nguyễn Việt Lâm]

1.

\(\left\{{}\begin{matrix}2x+3y=-2\\4x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-2\\-12x-3y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-10x=-5\\y=1-4x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1-4x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-1\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

2.

a.

\(P=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{6}{3\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)

\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{6}{\sqrt{x}+2}\)

\(=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{6}\)

\(=\dfrac{-1}{\sqrt{x}-2}\)

b.

\(Q=\left(-\sqrt{x}-1\right).P=-\left(\sqrt{x}+1\right).\dfrac{-1}{\sqrt{x}-2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(Q=\dfrac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}-2}\)

Q nguyên khi \(\dfrac{3}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{1;3;5\right\}\)

\(\Rightarrow x\in\left\{1;9;25\right\}\)