Question

Cho Q= \(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
a, Rút gọn Q
b, Tính Q khi x=\(\dfrac{2}{3-\sqrt{5}}\)
c, Tìm x thuộc Z để Q thuộc Z
d, Tìm x để Q <1
e, Tìm GTNN của \(\dfrac{1}{Q}\)

lm ơn giúp mik vs ạ !!!!

Answer 1

\(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

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

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

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