Question

P=\(\dfrac{\sqrt{x}+4}{1-7\sqrt{x}}+\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{24\sqrt{x}}{7x+6\sqrt{x}-1}\)

1. Tìm điều kiện để P có nghĩa và rút gọn P

2. Tìm x sao cho P >-6

3. Tìm x\(\varepsilon\)Z để PϵZ

Answer 1

1) \(=\dfrac{\left(\sqrt{x}+4\right)\left(\sqrt{x}+1\right)}{\left(1-7\sqrt{x}\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(7\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}+\dfrac{24\sqrt{x}}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}\)
\(=\dfrac{2\left(3\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}=\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}\)
2) \(P>-6=>\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0< =>\dfrac{8\left(\sqrt{x}-1\right)}{7\sqrt{x}-1}>0< =>\dfrac{\sqrt{x}-1}{7\sqrt{x}-1}>0\)
\(\left\{{}\begin{matrix}\sqrt{x}-1>0\\7\sqrt{x}-1>0\end{matrix}\right.< =>\left\{{}\begin{matrix}x>1\\x>\dfrac{1}{49}\end{matrix}\right.=>x>1\)