rút gọn : √(4+2√3) - √(13-4√3)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=2\sqrt{2}+1-3=2\sqrt{2}-2\)
b: \(=\sqrt{3}+1-2\sqrt{3}-1=-\sqrt{3}\)
c: \(=2-\sqrt{3}+\sqrt{3}-1=1\)
\(\frac{2^{13}\cdot3^7}{4^7\cdot9^3}=\frac{2^{13}\cdot3^7}{\left[2^2\right]^7\cdot\left[3^2\right]^3}=\frac{2^{13}\cdot3^7}{2^{14}\cdot3^6}=2^{-1}\cdot3=\frac{1}{2}\cdot3=\frac{3}{2}\)
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{8}{25-3}=\frac{8}{22}=\frac{4}{11}\)
\(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}=\frac{5-9}{14}=\frac{-4}{14}=\frac{-2}{7}\)
\(\frac{-2}{7}=\frac{-22}{77}\)
\(\frac{4}{11}=\frac{28}{77}\)
\(\sqrt{13-4\sqrt{3}}\)
\(=\sqrt{12-2.\sqrt{4}.\sqrt{3}+1}\)
\(=\sqrt{\sqrt{12^2}-2.\sqrt{1}.\sqrt{12}+\sqrt{1^2}}\)
\(=\sqrt{\left(\sqrt{12}-1\right)^2}\)
\(=\left|\sqrt{12}-1\right|\)
\(=\sqrt{12}-1\)
a: \(=\sqrt{5}-1\)
b: \(=\sqrt{2}-1\)
c: \(=\sqrt{3}+1\)
d: \(=\sqrt{13}+1\)
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{13-4\sqrt{3}}\)
\(A=\sqrt{3+2\sqrt{3}+1}-\sqrt{12-4\sqrt{3}+1}\)
\(A=\sqrt{\left(\sqrt{3}\right)^2+2\times\sqrt{3}\times1+1^2}-\sqrt{\left(2\sqrt{3}\right)^2-2\times2\sqrt{3}\times1+1^2}\)
\(A=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(A=\sqrt{3}+1-2\sqrt{3}+1\)
\(A=2-\sqrt{3}\)