Question

Cho A = \(\dfrac{2\sqrt{x}}{3+\sqrt{x}}\) : B = \(\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\)

1, Tính A khi x = \(\sqrt{10-\sqrt{19}}.\sqrt{10+\sqrt{19}}\)

2, Tính B khi x =\(\dfrac{1}{\sqrt{3}-2}+\sqrt{3}+66\)

Answer 1

1: \(x=\sqrt{100-19}=\sqrt{81}=9\)

Thay x=9 vào A, ta được:

\(A=\dfrac{2\cdot3}{3+3}=\dfrac{6}{6}=1\)

2: \(x=-2-\sqrt{3}+\sqrt{3}+66=64\)

Thay x=64 vào B, ta được:

\(B=\dfrac{8-2}{8-3}=\dfrac{6}{5}\)

Answer 2

1.

\(x=\sqrt{10-\sqrt{19}}.\sqrt{10+\sqrt{19}}\)

\(x=\sqrt{\left(10-\sqrt{19}\right)\left(10+\sqrt{19}\right)}\)

\(x=\sqrt{100-19}\)

\(x=\sqrt{81}=9\)

\(A=\dfrac{2\sqrt{x}}{3+\sqrt{x}}=\dfrac{2\sqrt{9}}{3+\sqrt{9}}=\dfrac{6}{6}=1\)

2.

\(x=\dfrac{1}{\sqrt{3}-2}+\sqrt{3}+66\)

\(x=\dfrac{1+\left(\sqrt{3}+66\right)\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\)

\(x=\dfrac{1+3-2\sqrt{3}+66\sqrt{3}-132}{\sqrt{3}-2}\)

\(x=\dfrac{64\sqrt{3}-128}{\sqrt{3}-2}\)

\(x=\dfrac{64\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\)

\(x=64\)

\(B=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}=\dfrac{\sqrt{64}-2}{\sqrt{64}-3}=\dfrac{8-2}{8-3}=\dfrac{6}{5}\)