Bài 2:
a: \(A=\sqrt{15a^2-8a\sqrt{15}+16}\)
\(=\sqrt{\left(a\sqrt{15}\right)^2-2\cdot a\sqrt{15}\cdot4+4^2}\)
\(=\sqrt{\left(a\sqrt{15}-4\right)^2}=\left|a\sqrt{15}-4\right|\)
\(a=\sqrt{\frac35}+\sqrt{\frac53}\)
\(=\frac{\sqrt{15}}{5}+\frac{\sqrt{15}}{3}=\frac{8\sqrt{15}}{15}\)
\(A=\left|a\sqrt{15}-4\right|\)
\(=\left|\sqrt{15}\cdot\frac{8\sqrt{15}}{15}-4\right|=\left|8-4\right|=4\)
b: \(B=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
\(=\sqrt{x^2-1+2\cdot\sqrt{x^2-1}\cdot1+1}-\sqrt{x^2-1-2\cdot\sqrt{x^2-1}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}-\sqrt{\left(\sqrt{x^2-1}-1\right)^2}=\sqrt{x^2-1}+1-\left|\sqrt{x^2-1}-1\right|\)
\(x=\sqrt5\)
=>\(x^2=5\)
=>\(x^2-1=5-1=4\)
\(B=\sqrt{x^2-1}+1-\left|\sqrt{x^2-1}-1\right|\)
=2+1-|2-1|
=2+1-1
=2
Bài 1:
\(A=\sqrt{4-\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt6\right)\)
\(=\sqrt{8-2\sqrt{15}}\left(\sqrt5-\sqrt3\right)\cdot\left(4+\sqrt{15}\right)\)
\(=\sqrt{\left(\sqrt5-\sqrt3\right)^2}\cdot\left(\sqrt5-\sqrt3\right)\left(4+\sqrt{15}\right)\)
\(=\left(\sqrt5-\sqrt3\right)\left(\sqrt5-\sqrt3\right)\left(4+\sqrt{15}\right)=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2\)
