Question

[Lê Trần Thuỳ Dương]

Giúp mik với mn

Answer 1

[l҉o҉n҉g҉ d҉z҉]

\(P=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\cdot\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\frac{x-1}{x}\)

Để P > 1/2 thì \(\frac{x-1}{x}>\frac{1}{2}\Leftrightarrow\frac{x-1}{x}-\frac{1}{2}>0\Leftrightarrow\frac{2x-2-x}{2x}>0\Leftrightarrow\frac{x-2}{2x}>0\)(1)

Vì x > 0 nên 2x > 0 nên (1) <=> x - 2 > 0 <=> x > 2