Đặt A = \(\sqrt{2004}+\sqrt{2006}\)
B = \(2\sqrt{2005}\)
Ta có : \(A^2=\left(\sqrt{2004}+\sqrt{2006}\right)^2\)
\(=\sqrt{2004}^2+2\sqrt{2004.2006}+\sqrt{2006}^2\)
\(=2004+2\sqrt{\left(2005-1\right)\left(2005+1\right)}+2006\)
\(=4010+2\sqrt{2005^2-1}\)
\(B^2=2.2005+2\sqrt{2005}^2=8020\)
\(\Rightarrow A^2< B^2\)
\(\Rightarrow A< B\)
so sánh
\(\sqrt{2004}-\sqrt{2003}và\sqrt{2006}-\sqrt{2005}\)
Tính: a, \(\sqrt{2006+2\sqrt{2005}}-\sqrt{2006-2\sqrt{2005}}\)
b, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
Rút gọn:
a) \(A=\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+... +\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2006\sqrt{2005}+2005\sqrt{2006}}+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\)
a) Chứng minh rằng với a > 0, b > 0 thì \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
b) So sánh \(\sqrt{2004+2005}\) với \(\sqrt{2004}+\sqrt{2005}\)
Chứng minh : \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...+\dfrac{1}{2005\sqrt{2004}}< 2\)
CMR 2004 < 1+ \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{1006009}}< 2005\)
CMR: \(\dfrac{1}{2}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+....+\dfrac{1}{2005\sqrt{2004}}< 2\)
\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+....+\dfrac{1}{\sqrt{2005}+\sqrt{2006}}\)
Giải chi tiết dùm mình hem
giải pt = cách đặt ẩn phụ
a) \(x^2+\sqrt{x+2006}=2006\)
b) \(x=\left(2004+\sqrt{x}\right)\left(1-\sqrt{1-\sqrt{x}}\right)^2\)
Làm chi tiết giúp mk vs
