Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[Edogawa Conan]

Đk: x \(\ge\)0 và x \(\ne\)1

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

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

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

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

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

b) Với x \(\ge\)0 và x \(\ne\)1 (1), ta có:

\(P< \frac{15}{4}\) <=> \(\frac{3\sqrt{x}+8}{\sqrt{x}+2}< \frac{15}{4}\) <=> \(12\sqrt{x}+32< 15\sqrt{x}+30\)

<=> \(3\sqrt{x}>2\) <=> \(\sqrt{x}>\frac{2}{3}\) <=> \(x>\frac{4}{9}\)(2)

Từ (1) và (2) => \(x>\frac{4}{9}\) và x \(\ne\)1