\(C=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
\(C^2=\left(\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\right)^2\)
\(C^2=x^2+2\sqrt{x^2-1}-2\sqrt{\left(x^2+2\sqrt{x^2-1}\right)\left(x^2-2\sqrt{x^2-1}\right)}+x^2-2\sqrt{x^2-1}\)
\(C^2=2x^2-2\sqrt{x^4-2x^2\sqrt{x^2-1}+2x^2\sqrt{x^2-1}-\left(2\sqrt{x^2-1}\right)^2}\)
\(C^2=2x^2-2\sqrt{x^4-4\left(x^2-1\right)}\)
\(C^2=2x^2-2\sqrt{x^4-4x^2+4}\)
\(C=\sqrt{2x^2-2\sqrt{x^4-4x^2+4}}\)
Thay: \(x=\sqrt{5}\) vào C, ta có:
\(C=\sqrt{2\sqrt{5}^2-2\sqrt{\sqrt{5}^4-4\sqrt{5}^2+4}}\)
\(C=\sqrt{10-2\sqrt{25-20+4}}\)
\(C=\sqrt{10-2\sqrt{9}}\)
\(C=\sqrt{10-6}\)
\(C=\orbr{\begin{cases}-2\\2\end{cases}}\)
Mà theo bài ra: \(\sqrt{x^2+2\sqrt{x^2-1}}>\sqrt{x^2-2\sqrt{x^2-1}}\)
\(\Rightarrow\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}>0\)
\(\Rightarrow C=2\)
Đề câu a là \(4\sqrt{5}a\) hay \(4\sqrt{5a}\) . Thấy \(4\sqrt{5}a\) đúng hơn
\(A=\sqrt{5a^2-4\sqrt{5}a+4}\)
\(A=\sqrt{\left(\sqrt{5}a-2\right)^2}\)
\(A=\sqrt{5}a-2\)
Thay \(a=\sqrt{5}+\frac{1}{\sqrt{5}}\) vào A, có:
\(A=\sqrt{5}\left(\sqrt{5}+\frac{1}{\sqrt{5}}\right)-2\)
\(A=5+1-2\)
\(A=4\)
\(B=\sqrt{15a^2-8a\sqrt{15}+16}\)
\(B=\sqrt{\left(\sqrt{15}a-4\right)^2}\)
\(B=\sqrt{15}a-4\)
Thay \(a=\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) vào B, có:
\(B=\sqrt{15}\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)-4\)
\(B=\sqrt{9}+\sqrt{25}-4\)
\(B=3+5-4\)
\(B=4\)
