Question

cho Q= (\(\dfrac{1}{x-\sqrt{x}}-\dfrac{1}{\sqrt[]{x}-1}\)) \(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)

a, tìm đkxđ, rút gọn Q

b, tính Q khi x= 9

c, tính x \(\left|Q\right|>-Q\)

d, tính x để Q\(\sqrt{Q}\)

Answer 1

a , Thu gọn :

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

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

\(Q=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)\)

\(Q=\dfrac{-1.\left(\sqrt{x}-1\right)}{\sqrt{x}}\)

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

b , Với x = 9 ta có :

\(Q=\dfrac{1-\sqrt{9}}{\sqrt{9}}=\dfrac{1-3}{3}=-\dfrac{2}{3}\)

a