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\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{23}+\sqrt{25}}\)
\(2A=\frac{2}{\sqrt{3}+\sqrt{1}}+\frac{2}{\sqrt{5}+\sqrt{3}}+...+\frac{2}{\sqrt{25}+\sqrt{23}}\)\(2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{\left(\sqrt{3}+\sqrt{1}\right)\left(\sqrt{3}-\sqrt{1}\right)}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{\left(\sqrt{25}+\sqrt{23}\right)\left(\sqrt{25}-\sqrt{23}\right)}\)
\(2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{2}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{2}\)
\(2A=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\sqrt{25}-\sqrt{23}\)
\(2A=\sqrt{25}-\sqrt{1}\)
\(2A=4\)
\(A=2\)
\(b,\) Ta có:
\(\dfrac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\\ =\dfrac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}\\ =\dfrac{\sqrt{n}}{\sqrt{n}.\sqrt{n-1}}-\dfrac{\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}\\ =\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\)
Thay:
\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)
\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(...\)
\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)
\(A=-\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+....-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}\)
\(A=\sqrt{9}-\sqrt{1}=3-1=2\)
a ) \(\dfrac{2}{\sqrt{3}-1}\) - \(\dfrac{2}{\sqrt{3}+1}\) = \(\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
= \(\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{3-1}\) = \(\dfrac{4}{2}\) = 2
b) \(\dfrac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}\) - \(\dfrac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{5\left(2\sqrt{5}-3\sqrt{2}\right)-5\left(2\sqrt{5}+3\sqrt{2}\right)}{12\left(2\sqrt{5}+3\sqrt{2}\right)\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{10\sqrt{5}-15\sqrt{2}-10\sqrt{5}-15\sqrt{2}}{12\left(20-18\right)}\)
= \(\dfrac{-30\sqrt{2}}{24}\) = \(\dfrac{-15\sqrt{2}}{12}\) = \(\dfrac{-5\sqrt{2}}{4}\)
c) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) +\(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\) = \(\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
= \(\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\) = \(\dfrac{60}{20}\) = 3
d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}+1}\)
= \(\dfrac{\sqrt{3}}{\sqrt{2}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{2}+1}\) = \(\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
= \(\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}}{2-1}\) = \(2\sqrt{3}\)
\(P=\frac{1}{\sqrt{1}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{\sqrt{5}-\sqrt{1}}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+...+\frac{\sqrt{2005}-\sqrt{2001}}{4}\)
\(=\frac{\sqrt{2005}-\sqrt{1}}{4}=\frac{\sqrt{2005}-1}{4}\)
ĐKXĐ:...
\(A=\left(\frac{\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a}+1}\right).\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}=\frac{1}{a}\)
\(C=\left(\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(=\left(\frac{\left(\sqrt{x}+1\right)}{-\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)}.\frac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)}\)
\(=\left(-1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\sqrt{x}=\left(\frac{-x-\sqrt{x}-1+x+\sqrt{x}}{x+\sqrt{x}+1}\right)\sqrt{x}=\frac{-\sqrt{x}}{x+\sqrt{x}+1}\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
\(P=\frac{\sqrt{5}-1}{5-1}+\frac{\sqrt{9}-\sqrt{5}}{9-5}+...+\frac{\sqrt{2005}-\sqrt{2001}}{2005-2001}=\frac{-1+\sqrt{2005}}{4}\)
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)
\(\Rightarrow A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{2}}\right)\left(4-\sqrt{10+2\sqrt{2}}\right)}+4-\sqrt{10+2\sqrt{5}}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5.1}+1}=8+2\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
\(B=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)
\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(=-\frac{1}{4}\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+....+\sqrt{2001}-\sqrt{2005}\right)\)
\(=-\frac{1}{4}\left(1-\sqrt{2005}\right)\)
\(=10,94430659\)
\(\text{Lm hơi vắn tắt thông cảm nha!!}\)
Lười chép đề quá, thông cảm ._.
P = \(1-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2005}}\)
P = \(1-\frac{1}{\sqrt{2005}}\)
P = \(\frac{1}{1+\sqrt{2005}}\)
Chúc bạn học tốt!