
\(\sqrt{4+2\sqrt{3}}-\sqrt{5+2\sqrt{6}}+\sqrt{2}\)
b) Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. Bài làm: a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\) \(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\) \(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\) \(A=4+2\sqrt{3}+5\sqrt{3}-1\) \(A=3+7\sqrt{3}\) b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\) \(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\) \(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\) \(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\) \(A=2\) Phần b mình viết nhầm tên thành A, bn sửa thành B nhé c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\) \(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\) \(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\) \(C=\sqrt{3}-1-2-\sqrt{3}\) \(C=-3\) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+\sqrt{48}}}}\) \(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2-\sqrt{3}\right)^2}}}\) \(=\sqrt{5\sqrt{3}+5\sqrt{48-20+10\sqrt{3}}}\) \(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\) \(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\) = 5 \(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\) \(=\dfrac{\sqrt{3}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\) \(=\dfrac{\sqrt{3}-\sqrt{2}-\sqrt{3}+3+\sqrt{2}}{\sqrt{5}+1+\sqrt{2}-\sqrt{2}-\sqrt{5}}\) \(=3\) a: \(=6\sqrt{2}-12\sqrt{3}-10\sqrt{2}+12\sqrt{3}=-4\sqrt{2}\) b: \(=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{4-3}=1\)
