Chương I - Căn bậc hai. Căn bậc ba

TT

Trục căn thức và thực hiện phép tính:

\(N=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

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

\(Q=\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)

H24
7 tháng 8 2022 lúc 16:18

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

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