ND

Tính A = ( \(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}\))\(^2\) - (\(\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\))\(^2\)

H9
17 tháng 9 2023 lúc 9:55

\(A=\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}\right)^2-\left(\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)^2\)

\(A=\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}+\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)

\(A=\left[\dfrac{2\left(2-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\dfrac{2\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\right]\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)

\(A=\left[\dfrac{2\left(2\sqrt{3}+2-\sqrt{15}-\sqrt{5}\right)+2\left(2\sqrt{3}-2+3-\sqrt{3}\right)}{3-1}\right]\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)

\(A=\left(2\sqrt{3}+2-\sqrt{15}-\sqrt{5}+2\sqrt{3}+1-\sqrt{3}\right)\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)

\(A=\left(3\sqrt{3}+3-\sqrt{15}-\sqrt{5}\right)\left[\dfrac{2\left(2\sqrt{3}+2-\sqrt{15}-\sqrt{5}-2\sqrt{3}+2-3+\sqrt{3}\right)}{2}\right]\)

\(A=\left(3\sqrt{3}+3-\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1-\sqrt{15}-\sqrt{5}\right)\)

\(A=\left(\sqrt{3}+1\right)^2\left(3-\sqrt{5}\right)\left(1-\sqrt{5}\right)\)

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