Question

Cho b.thức: P= \(\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)

a) Rút gọn P

b)Tính giá trị của A với x=\(14-6\sqrt{5}\)

Answer 1

ĐKXĐ : \(x\ne9\)

a)

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

= \(\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

= \(\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+1\right)}\)

= \(\dfrac{\left(x\sqrt{x}-3x\right)+\left(8\sqrt{x}-24\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+1\right)}\) = \(\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-3\right).\left(x+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

=\(\dfrac{x+8}{\sqrt{x}+1}\)

Answer 2

b)

thay x = \(14-6\sqrt{5}\) ( thỏa mãn ĐKXĐ ) vào biểu thức P ta được :

P = \(\dfrac{14-6\sqrt{5}+8}{\sqrt{14-6\sqrt{5}}+1}=\dfrac{22-6\sqrt{5}}{\sqrt{\left(3-\sqrt{5}\right)^2}+1}\)

= \(\dfrac{22-6\sqrt{5}}{4-\sqrt{5}}=\dfrac{58-2\sqrt{5}}{11}\)

Vậy...