Question

Cho Q=\(\dfrac{x+\sqrt{x}}{\left(x-\sqrt{x}+1\right)^2}\);(với x≥0;x≠1)

Tìm x để Q<1

Answer 1

Để Q<1 thì Q-1<0

\(\Leftrightarrow\dfrac{x+\sqrt{x}}{\left(x-\sqrt{x}+1\right)^2}-\dfrac{\left(x-\sqrt{x}+1\right)^2}{\left(x-\sqrt{x}+1\right)^2}< 0\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}-\left(x^2+x+1-2x\sqrt{x}+2x-2\sqrt{x}\right)}{\left(x-\sqrt{x}+1\right)^2}< 0\)

mà \(\left(x-\sqrt{x}+1\right)^2>0\forall x\) thỏa mãn ĐKXĐ

nên \(x+\sqrt{x}-x^2-x-1+2x\sqrt{x}-2x+2\sqrt{x}< 0\)

\(\Leftrightarrow-x^2+2x\sqrt{x}-2x+3\sqrt{x}-1< 0\)

\(\Leftrightarrow-\left(x^2-2x\sqrt{x}+2x-3\sqrt{x}+1\right)< 0\)

\(\Leftrightarrow x^2-2x\sqrt{x}+2x-3\sqrt{x}+1>0\)

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