Question

don gian cac bieu thuc

can (2 + can 3) - can(2 - can 3)

Answer 1

đặt A=\(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

=>\(\sqrt{2}A=\sqrt{2}\sqrt{2+\sqrt{3}}-\sqrt{2}\sqrt{2-\sqrt{3}}\)

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

=\(\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}\)

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

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

=>A=\(\frac{2}{\sqrt{2}}=\frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}=\sqrt{2}\)

vậy \(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\sqrt{2}\)

Answer 2

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