Question

1, cho E =\(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\right)-\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)

a, rút gọn E (ĐKXĐ: x>=0 ; x khác 1)

b tìm x để E=\(-\dfrac{1}{7}\)

c, tìm x để E >0

2 cho hàm số y=x² có đồ thị là (P) và đường thẳng y=x-m+1(d)

tìm m để (d) cắt (P) tại 2 điểm pb ở bên phải trục tung sao cho x₂=2x₁

Answer 1

1/ ĐKXĐ: \(x\ge0,x\ne1\)

\(E=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\right)-\left(1-\dfrac{\sqrt{x}}{x+1}\right)\)

= \(\left[\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right]-\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)

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

= \(\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{x+1+\sqrt{x}}{x+1}\)

= \(\dfrac{1-\sqrt{x}}{x+1}-\dfrac{x+1+\sqrt{x}}{x+1}\)

= \(\dfrac{1-\sqrt{x}-x-1-\sqrt{x}}{x+1}=\dfrac{-x-2\sqrt{x}}{x+1}\)

b/ Với \(x\ge0,x\ne1\)

Để \(E=-\dfrac{1}{7}\Leftrightarrow\dfrac{-x-2\sqrt{x}}{x+1}=-\dfrac{1}{7}\)

\(\Leftrightarrow-7x-14\sqrt{x}+x+1=0\)

\(\Leftrightarrow-6x-14\sqrt{x}+1=0\)

\(\Leftrightarrow\left(6\sqrt{x}+7-\sqrt{55}\right)\left(6\sqrt{x}+7+\sqrt{55}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}6\sqrt{x}+7-\sqrt{55}=0\\6\sqrt{x}+7+\sqrt{55}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{-7+\sqrt{55}}{6}\\\sqrt{x}=\dfrac{-7-\sqrt{55}}{6}\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{52-7\sqrt{55}}{18}\)

Vậy để \(E=-\dfrac{1}{7}\) thì \(x=\dfrac{52-7\sqrt{55}}{18}\)