Tìm x bt:
\(\sqrt{x^2+2x+1}\) = -x
Rút gọn:
a, \(\sqrt{\left(4-\sqrt{17}\right)}^2\) - \(\sqrt{17}\)
b, \(\sqrt{\left(5-2\sqrt{3}\right)^2}\) - \(2\sqrt{3}\)
BT: Tính
a, \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
b,\(\left(3-\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\cdot\sqrt{3-\sqrt{5}}\)
c,\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
(1)\(\frac{\sqrt{6+4\sqrt{2}}}{\sqrt{2}}\) (2)\(\frac{\sqrt{3-\sqrt{5}}}{\sqrt{0.5}}\) (3)\(\left(\sqrt{2}-1\right)^2\) (4)\(\left(3-2\sqrt{2}\right).\left(3+2\sqrt{2}\right)\) (5)\(\sqrt{\left(2-\sqrt{3}\right)}^2-\sqrt{\left(1-\sqrt{3}\right)}^2\) (6)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)}^2\) (7)\(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\) (8)\(\sqrt{3-2\sqrt{2}}\) (9)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\) (10)\(\sqrt{2020+2\sqrt{2019}}-\sqrt{2020-2\sqrt{2019}}\) (11)\(\sqrt{7+2\sqrt{12}}\) Các bạn giúp mình với ,Mình xin cảm ơn trước
Tính:
a) \(A=\sqrt{8-2\sqrt{15}}\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
b) \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
c) \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}+}\sqrt{3}\right):\sqrt{3}\)
d) \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
a, \(A=\left(\sqrt{2}+1\right)[\left(\sqrt{2}\right)^2+1][(\sqrt{2})^4+1][\left(\sqrt{2}\right)^8+1][1\left(\sqrt{2}\right)^{16}+1]\)
b, \(B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+1\sqrt{2020}}\)
c,\(C=^3\sqrt[]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\)
Rút gọn biểu thức:
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}\)
\(B=\left(\frac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\frac{4}{1+\sqrt{5}}+4\right)\)
\(C=\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right):\left(1:\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(D=2\sqrt{50}-\frac{1}{\sqrt{2}-1}+4\sqrt{\frac{9}{2}}-\sqrt{3-2\sqrt{2}}\)
1 )\(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
2) \(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)
3)\(\sqrt{4-2\sqrt{3}}-\dfrac{3+\sqrt{3}}{\sqrt{3}-1}+\dfrac{2}{\sqrt{3}-1}\)
\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{4+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
1. Rút gọn \(A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\)
2. Tính \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
3.Tính \(C=\frac{\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\cdot\left(3+\sqrt{5}\right)}{\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)