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a/ \(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2=2\left(sin^2\alpha+cos^2\alpha\right)=2\)
b/ \(B=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)-\left(1+cotg^2\alpha\right)\left(1-cos^2\alpha\right)\)
\(=\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)\left(1-sin^2\alpha\right)-\left(1+\frac{cos^2\alpha}{sin^2\alpha}\right)\left(1-cos^2\alpha\right)\)
\(=\frac{1}{cos^2\alpha}.cos^2\alpha-\frac{1}{sin^2\alpha}.sin^2\alpha=1-1=0\)
a) \(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=1-\cos^2\alpha=\sin^2\alpha\)
Ta có: \(\dfrac{1}{2+\sqrt{3}}+\sqrt{3}\)
\(=2-\sqrt{3}+\sqrt{3}\)
=2
\(\dfrac{6}{\sqrt{2}-\sqrt{3}+3}\)
\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{\left(\sqrt{2}-\sqrt{3}\right)^2-9}\)
\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\)
\(=\dfrac{-3\left(\sqrt{2}-\sqrt{3}-3\right)}{2+\sqrt{6}}=\dfrac{-3\left(\sqrt{6}-2\right)\left(\sqrt{2}-\sqrt{3}-3\right)}{2}\)
Ta có: \(\dfrac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}}{\sqrt{3}-1}\)
\(=\sqrt{2}\)