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Ta có công thức tổng quát
\(\dfrac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)Vậy \(P=\dfrac{1}{\sqrt{2}.1+\sqrt{1}.2}+\dfrac{1}{\sqrt{3}.2+\sqrt{2}.3}+...+\dfrac{1}{\sqrt{100}.99+\sqrt{99}.100}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(1.A=\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right).\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\left(\dfrac{3+\sqrt{5}}{9-5}-\dfrac{3-\sqrt{5}}{9-5}\right).\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\dfrac{2\sqrt{5}}{4}.\sqrt{5}=\dfrac{5}{2}\) \(2.B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}=\sqrt{100}-1\)
\(3.C=\sqrt[3]{7+5\sqrt{2}}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.2.1+3.\sqrt{2}.1+1}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.2.1+3.\sqrt{2}.1-1}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\) \(4.Sai-đề\) ???
Sorry và cám ơn bạn.
4.\(\sqrt[3]{9+4\sqrt{5}}\) + \(\sqrt[3]{9-4\sqrt{5}}\)
Chứng minh đẳng thức phụ:
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)
\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}\)
\(\Rightarrow\) Với \(a+b+c=0\). Ta có: \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
\(\Leftrightarrow\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\)với \(a+b+c=0\)
Ta có:
\(S=\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+.....+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\)
Áp dụng đẳng thức phụ trên:
\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{1^2}+\dfrac{1}{\left(-2\right)^2}}=1+1-\dfrac{1}{2}\left(>0\right)\)
\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{\left(-3\right)^2}}=1+\dfrac{1}{2}-\dfrac{1}{3}\left(>0\right)\)
\(\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{\left(-4\right)^2}}=1+\dfrac{1}{3}-\dfrac{1}{4}\left(>0\right)\)
\(.................\)
\(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{99^2}+\dfrac{1}{\left(-100\right)^2}}=1+\dfrac{1}{99}-\dfrac{1}{100}\)
Cộng vế với vế các đẳng thức trên, ta có:
\(S=\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+........+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\)
\(=1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+............+1+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=99+1-\dfrac{1}{100}=99+\dfrac{99}{100}=99\dfrac{99}{100}\)
Chứng minh kiểu khác :v
\(\forall n\in\)N*, ta có:
\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{\left[n.\left(n+1\right)\right]^2+2n\left(n+1\right)+1}{\left[n\left(n+1\right)\right]^2}}\)
\(=\sqrt{\dfrac{\left[n\left(n+1\right)+1\right]^2}{\left[n\left(n+1\right)\right]^2}}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)
Việc còn lại là áp dụng vào bài thoy :v
\(a.A=\sqrt{1+\dfrac{1}{x^2}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{x}\right)^2-\dfrac{2}{x}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\left(\dfrac{x+1}{x}\right)^2-2.\dfrac{x+1}{x}.\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{x}-\dfrac{1}{x+1}\right)^2}=\left|x+\dfrac{1}{x}+\dfrac{1}{x+1}\right|\)
\(b.\) Áp dụng điều đã CM ở câu a , ta có :
\(B=\sqrt{1+\dfrac{1}{1^1}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=100-\dfrac{1}{100}=\)
Bạn thử tham khảo link này nha: https://olm.vn/hoi-dap/question/1294056.html
Bài 50:
\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{1}{3\sqrt{20}}=\dfrac{1}{6\sqrt{5}}=\dfrac{\sqrt{5}}{30}\)
\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
=(\(\dfrac{99}{2}+1+\dfrac{98}{3}+1+...+\dfrac{1}{100}+1\)):(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\)) -2
=(\(\dfrac{101}{2}+\dfrac{101}{3}+...\dfrac{101}{100}\)):(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\)) -2
=101(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\)):(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\))-2
=101 -2 =99
-_-