Câu 2:
Gọi số học sinh của lớp 9A là x(bạn)(Điều kiện: \(x\in Z^+\))
Theo đề, ta có phương trình:
\(\dfrac{300}{x-10}=\dfrac{300}{x}+5\)
\(\Leftrightarrow\dfrac{300x}{x\left(x-10\right)}=\dfrac{300\left(x-10\right)}{x\left(x-10\right)}+\dfrac{5x\left(x-10\right)}{x\left(x-10\right)}\)
Suy ra: \(5x^2-50x+300x-3000-300x=0\)
\(\Leftrightarrow x^2-10x-600=0\)
\(\Leftrightarrow x^2-30x+20x-600=0\)
\(\Leftrightarrow\left(x-30\right)\cdot\left(x+20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=30\left(nhận\right)\\x=-20\left(loại\right)\end{matrix}\right.\)
Vậy: Lớp 9A có 30 bạn
Câu 1:
1) Ta có: \(\left\{{}\begin{matrix}x+3y=4\\2x-3y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=3\\x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\3y=4-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy: (x,y)=(1;1)
2) Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}+2}-\dfrac{8}{x-4}\right):\dfrac{x+4}{x\sqrt{x}-4\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+4-4\left(\sqrt{x}-2\right)-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{x+4}\)
\(=\dfrac{x+4\sqrt{x}+4-4\sqrt{x}+8-8}{x+4}\cdot\sqrt{x}\)
\(=\dfrac{x+4}{x+4}\cdot\sqrt{x}=\sqrt{x}\)
