Question

Cho biết: \(D=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)

a, Rút gọn D.

b, Tính D khi x = \(\frac{1}{4}\).

c, Tìm x để D = 2.

d, Tìm x để D >3.

Answer 1

bài khá dễ nhưng làm thì mắc công quá

Answer 2

a) Rgọn

D= \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)

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

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

=\(\frac{\sqrt{x}-1}{\sqrt{x}-3}\)

b)thay x=\(\frac{1}{3}\)ta dc

\(\frac{\sqrt{\frac{1}{4}}-1}{\sqrt{\frac{1}{4}-3}}\)=\(\frac{\frac{1}{2}-1}{\frac{1}{2}-3}\)=\(\frac{1}{2}\)

c) Để D=2 => \(\frac{\sqrt{x}-1}{\sqrt{x}-3}=2\)

=> 2\(\sqrt{x}-2=\sqrt{x}-3\)

=> \(\sqrt{x}=-1\)

=> x =1