Violympic toán 9

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Ngô Hoàng Anh

Bài 1 :

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

a, Tìm ĐKXĐ

b, Rút gọn

c, Tìm x để T < 0

Help me !!!

Nguyễn Kim Chi
17 tháng 8 2019 lúc 9:05

a, ĐKXĐ: x>0 (1)

b,T= (\(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)+6\sqrt{x}-9\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}\)).(\(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\))+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)

= \(\left(\frac{x+3\sqrt{x}+2+6\sqrt{x}-9x-9\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)

= \(\left(\frac{2-8x}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)

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

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

c, Để T<0 \(\Leftrightarrow\)\(\frac{\sqrt{x}-1}{3}\) <0 \(\Leftrightarrow\) \(\sqrt{x}\)-1<0 \(\Leftrightarrow\) \(\sqrt{x}\)<1\(\Leftrightarrow\) x<1 mà do ĐK (1)

=> Để T<0 \(\Leftrightarrow\) 0<x<1