Question

Cho \(P=\dfrac{3x-\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

a) Rút gọn P .

b) Tính giá trị P khi x = 3+2\(\sqrt{2}\)

Answer 1

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

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

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

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

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b, x=\(3+2\sqrt{2}\)=\(2+2\sqrt{2}+1\)=\(\left(\sqrt{2}+1\right)^2\)

Thay x vào P ta có:

\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}-2}{\sqrt{\left(\sqrt{2}+1\right)^2}+2}\)=\(\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1+2}\)=\(\dfrac{-5+4\sqrt{2}}{7}\)

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