Question

Bài 1 Cho biểu thức

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

1 Rút gọn H

2 Tìm x để H < 0

b) Q = \(\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}+1\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-1\right)\)

1 Rút gọn Q

2 Tìm x để Q < 1

GIÚP MÌNH VS CẦN GẤP LẮM NHA !!!!!!

Answer 1

B1:

a)

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

b)

\(H< 0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}< 0\)

vì \(\sqrt{x}\ge0\)

nên \(\sqrt{x}-1< 0\\ \sqrt{x}< 1\Rightarrow x< 1\)

vậy khi x<1 thì H < 0

Answer 2

b)

1.

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

2.

\(Q< 1\Leftrightarrow x-1< 1\Leftrightarrow x< 2\)

vậy khi x< 2 thì Q<1