Question

2.4 Rút gọn biểu thức

\(a,\dfrac{3-\sqrt{x}}{x-9}\) ( vs x ≥ 0, x≠ 9)

b, \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}\)( vs x ≥ 0 ; x ≠ 9)

c, \(6-2x-\sqrt{9-6x+x^2}\left(x< 3\right)\)

Answer 1

a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x+3}}\)(\(x\ge0,x\ne9\))

b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\sqrt{x}-2\left(x\ge0,x\ne9\right)\)

 

Answer 2

a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x}+3}\)

b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)

c) \(6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-\left|3-x\right|\)

mà \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)

Answer 3

a,\(\dfrac{3-\sqrt{x}}{x-9}\)

=\(-\dfrac{3-\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

=\(-\dfrac{1}{3+\sqrt{x}}\)

Answer 4

c,\(6-2x-\sqrt{9-6x+x^2}\)

=\(6-2x-\sqrt{-\left(x-3\right)^2}\)

=\(6-2x-\left|-\left(x-3\right)\right|\)

=\(6-2x-\left(3-x\right)\)(x<3)

=3-x