Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\sqrt{6-4\sqrt2}+\sqrt{22-12\sqrt2}\)
\(=\sqrt{4-2\cdot2\cdot\sqrt2+2}+\sqrt{18-2\cdot3\sqrt2\cdot2+4}\)
\(=\sqrt{\left(2-\sqrt2\right)^2}+\sqrt{\left(3\sqrt2-2\right)^2}\)
\(=2-\sqrt2+3\sqrt2-2=2\sqrt2\)
b: \(\sqrt{\left(\sqrt3-\sqrt2\right)^2}+\sqrt2=\sqrt3-\sqrt2+\sqrt2=\sqrt3\)
c: \(3\sqrt5-\sqrt{\left(1-\sqrt5\right)^2}\)
\(=3\sqrt5-\left|1-\sqrt5\right|\)
\(=3\sqrt5-\left(\sqrt5-1\right)=2\sqrt5+1\)
d:Sửa đề: \(\sqrt{17-12\sqrt2}+\sqrt{6+4\sqrt2}\)
\(=\sqrt{9-2\cdot3\cdot2\sqrt2+8}+\sqrt{4+2\cdot2\cdot\sqrt2+2}\)
\(=\sqrt{\left(3-2\sqrt2\right)^2}+\sqrt{\left(2+\sqrt2\right)^2}=3-2\sqrt2+2+\sqrt2=5-\sqrt2\)
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
B. 2
\(\frac{\left(5\sqrt3+\sqrt{50}\right)\left(\sqrt5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt2}\)
\(=\frac{\left(5\sqrt3+5\sqrt2\right)\left(\sqrt5-2\sqrt6\right)}{5\sqrt3-5\sqrt2}=\frac{5\left(\sqrt3+\sqrt2\right)\left(\sqrt5-2\sqrt6\right)}{5\left(\sqrt3-\sqrt2\right)}\)
\(=\frac{\left(\sqrt3+\sqrt2\right)\left(\sqrt5-2\sqrt6\right)}{\left(\sqrt3-\sqrt2\right)}=\left(\sqrt3+\sqrt2\right)^2\cdot\left(\sqrt5-2\sqrt6\right)\)
\(=\left(5+2\sqrt6\right)\left(\sqrt5-2\sqrt6\right)=5\sqrt5-10\sqrt6+2\sqrt{30}-24\)
=>Không có câu nào đúng