Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(a=\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}=\frac{\sqrt{3}}{\sqrt{5}}+\frac{\sqrt{5}}{\sqrt{3}}=\frac{8\sqrt{15}}{15}\)
=> \(a^2=\frac{64}{15}\)
=> \(M=\sqrt{15a^2-8a\sqrt{15}+16}=\sqrt{15.\frac{64}{15}-8.\frac{8\sqrt{15}}{15}.\sqrt{15}+16}\)
\(M=\sqrt{64-64+16}=4\)
\(C=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
\(C^2=\left(\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\right)^2\)
\(C^2=x^2+2\sqrt{x^2-1}-2\sqrt{\left(x^2+2\sqrt{x^2-1}\right)\left(x^2-2\sqrt{x^2-1}\right)}+x^2-2\sqrt{x^2-1}\)
\(C^2=2x^2-2\sqrt{x^4-2x^2\sqrt{x^2-1}+2x^2\sqrt{x^2-1}-\left(2\sqrt{x^2-1}\right)^2}\)
\(C^2=2x^2-2\sqrt{x^4-4\left(x^2-1\right)}\)
\(C^2=2x^2-2\sqrt{x^4-4x^2+4}\)
\(C=\sqrt{2x^2-2\sqrt{x^4-4x^2+4}}\)
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: \(\sqrt{x^2+2\sqrt{x^2-1}}>\sqrt{x^2-2\sqrt{x^2-1}}\)
\(\Rightarrow\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}>0\)
\(\Rightarrow C=2\)
2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)
b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) thì
\(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)
\(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)
\(=\sqrt{4^2}\)
\(=4\)
Giải
Ta có: \(\sqrt{\dfrac{5}{3}}+\sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{5}}{\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{5}}=\dfrac{8}{\sqrt{15}}\)
Vậy M = \(\sqrt{15\left(\dfrac{8}{15}\right)^2-8.\dfrac{8}{\sqrt{15}}.\sqrt{15}+16}\)
= \(\sqrt{8^2-8^2+16}=\sqrt{16}=4\)
a) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0
a)\(\sqrt{\dfrac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\dfrac{\sqrt{b}-1}{\sqrt{a}+1}}\)
=\(\dfrac{\sqrt{\sqrt{a}-1}}{\sqrt{\sqrt{b}+1}}.\dfrac{\sqrt{\sqrt{a}+1}}{\sqrt{\sqrt{b}-1}}\)
=\(\sqrt{\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}}\)
=\(\sqrt{\dfrac{a-1}{b-1}}\)(1)
Thay a=7,25, b=3,25 vào(1)
=>=\(\sqrt{\dfrac{7,25-1}{3,25-1}}=\sqrt{\dfrac{6,25}{2,25}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\)
b) =\(\sqrt{\left(\sqrt{15}a+4\right)^2}=\sqrt{15}a+4\)
Thay a
=>\(\sqrt{15}.\left(\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{5}{3}}\right)+4=3+5+4=12\)
c)
=\(\sqrt{\left(a^2-1\right)+2\sqrt{a^2-1}+1}-\sqrt{\left(a^2-1\right)-2\sqrt{a^2-1}+1}\)
=\(\sqrt{\left(a^2-1+1\right)^2}-\sqrt{\left(a^2-1-1\right)^2}\)
=\(a^2-\left(a^2-2\right)\)
=\(2\)
\(M=\sqrt{15a^2-8a\sqrt{15}+16}=\sqrt{\left(\sqrt{15}a-4\right)^2}\)
\(=\sqrt{15}a-4=\sqrt{15}\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)-4\)
\(=\left(3+5\right)-4=4\)