Question

P=\(\dfrac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\) Rút gọn và tìm giá trị của x để P<\(\dfrac{15}{4}\)

Answer 1

\(P=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

P<15/4

=>P-15/4<0

=>\(\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}-\dfrac{15}{4}< 0\)
=>\(4\left(3\sqrt{x}+8\right)-15\sqrt{x}-30< 0\)

=>\(-3\sqrt{x}+2< 0\)

=>\(-3\sqrt{x}< -2\)

=>x>4/9