Không dùng máy tính hãy so sánh P với 20
\(P=\sqrt{102-2\sqrt{101}}+\sqrt{103+2\sqrt{101}}\)
\(P=\sqrt{101-2\sqrt{101}+1}+\sqrt{101+2\sqrt{101}+1+1}\)
\(=\sqrt{\left(\sqrt{101}-1\right)^2}+\sqrt{\left(\sqrt{101}+1\right)^2+1}>\sqrt{101}-1+\sqrt{101}+1=2\sqrt{101}>2.\sqrt{100}=2.10=20\)
=> P > 20
1) Cho A= \(\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{101}}+\dfrac{1}{\sqrt{102}}+\dfrac{1}{\sqrt{103}}+\dfrac{1}{\sqrt{104}}\)
và B= \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)....\left(1-\dfrac{1}{100^2}\right)\)
So sánh A và B.
Rút gọn A = \(\frac{1}{3+\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+....+\frac{1}{101\sqrt{103}+103\sqrt{101}}\)
đơn giản biểu thức \(\frac{1}{3+\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+...+\frac{1}{101\sqrt{103}+103\sqrt{101}}\)
help me !
tính S = \(\frac{1}{3+\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+\sqrt{5}7}+.....+\frac{1}{101\sqrt{103}+103\sqrt{101}}\text{ [}\)!
Xét biểu thức phụ : \(\frac{1}{\left(2n+3\right)\sqrt{2n+1}+\left(2n+1\right)\sqrt{2n+3}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)
\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}\)
\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{2\sqrt{2n+1}.\sqrt{2n+3}}=\frac{1}{2}\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\ge1\)
Áp dụng : \(S=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{101\sqrt{103}+103\sqrt{101}}\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{103}}\right)\)
Tất cả bằng 1 tin đi
help me !
tính S = \(\frac{1}{3+\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+\sqrt{5}7}+.....+\frac{1}{101\sqrt{103}+103\sqrt{101}}\text{Doumo arigatou}\)!
Tính
\(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{101\sqrt{100}+100\sqrt{101}}\)
Chứng minh rằng: \(\frac{1}{\sqrt{101}}+\frac{1}{\sqrt{102}}+\frac{1}{\sqrt{103}}+...+\frac{1}{\sqrt{256}}<12\)
Áp dụng
\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}<\frac{2}{\sqrt{n-1}+\sqrt{n}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\)
có phải không?
trời ơi mk mà lm đc chắc đi thi hsg thế giới mất !!!
Chứng minh rằng : \(\frac{1}{\sqrt{101}}+\frac{1}{\sqrt{102}}+\frac{1}{\sqrt{103}}+...+\frac{1}{\sqrt{256}}<12\)
