Bài 1: Căn bậc hai
Các câu hỏi tương tự
1.Chứng minh: \(\frac{1}{2\cdot\sqrt{1}}+\frac{1}{3\cdot\sqrt{2}}+\frac{1}{4\cdot\sqrt{3}}+...+\frac{1}{2012\cdot\sqrt{2011}}+\frac{1}{2013\cdot\sqrt{2012}}\)\(< 2\)
2.Chứng minh: A= \(\frac{1}{3\cdot\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\cdot\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{97\cdot\left(\sqrt{48}+\sqrt{49}\right)}\)\(< \frac{1}{2}\)
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. 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 \(C=\frac{\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\cdot\left(3+\sqrt{5}\right)}{\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)
giải phương trình :
\(\sqrt{x-2}+2\cdot\sqrt{x-3}+\sqrt{x+6+6\cdot\sqrt{x-3}}=4\)
Tính giá trị của A=\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}+2020\cdot\sqrt{2}\)
Tìm x biết: \(\sqrt{x+3+14\cdot\sqrt{x-1}}\sqrt{x+8-6\cdot\sqrt{x-1}}=5\)
\(\frac{2\cdot x^4-5\cdot x^3+2\cdot x^2-5\cdot x-30}{x^2+10\cdot x-15}\) với x=\(-\sqrt{5}\)
\(\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(\left(\sqrt{10}-\sqrt{2}\right)\cdot\sqrt{3+\sqrt{5}}\)