Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\dfrac{\sqrt[3]{4}+\sqrt[3]{2}+2}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\dfrac{\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{8}}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)
\(=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{2}+1+\sqrt[3]{4}\right)}{\sqrt[3]{2}+1+\sqrt[3]{4}}=\sqrt[3]{2}\)
b)\(\sqrt{3+\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}}=\sqrt{3+\sqrt{3}+\sqrt[3]{\left(1+\sqrt{3}\right)^3}}\)
\(=\sqrt{3+\sqrt{3}+1+\sqrt{3}}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(1+\sqrt{3}\right)^2}=1+\sqrt{3}\)
c)\(\dfrac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}=\dfrac{\left(1+\sqrt{3}\right)^2}{\sqrt[3]{\left(1+\sqrt{3}\right)^3}}=\dfrac{\left(1+\sqrt{3}\right)^2}{1+\sqrt{3}}\)=\(1+\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
Bài 3:
a: \(A=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{x-25}\)
\(=\dfrac{x-10\sqrt{x}+25}{x-25}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b: \(B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}=\dfrac{3}{\sqrt{x}+3}\)
2: \(=\sqrt{2}-1-\sqrt{2}=-1\)
3: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)
4: \(=1+\dfrac{2-\sqrt{3}}{2-\sqrt{3}}=1+1=2\)
Đặt \(B=\sqrt{4+\sqrt3}+\sqrt{4-\sqrt3}\)
=>\(B^2=4+\sqrt3+4-\sqrt3+2\cdot\sqrt{\left(4+\sqrt3\right)\left(4-\sqrt3\right)}=8+2\sqrt{13}\)
=>\(B=\sqrt{8+2\sqrt{13}}\)
\(A=\frac{\sqrt{4+\sqrt3}+\sqrt{4-\sqrt3}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt2}\)
\(=\sqrt{\frac{8+2\sqrt{13}}{4+\sqrt3}}+\sqrt{\left(5-\sqrt2\right)^2}=\sqrt2+5-\sqrt2=5\)