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\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\\ =\dfrac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\dfrac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+....+\dfrac{1}{\sqrt{24.25}\left(\sqrt{25}+\sqrt{24}\right)}\\ =\dfrac{\sqrt{2}-1}{\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}.\sqrt{3}}+...+\dfrac{\sqrt{25}-\sqrt{24}}{\sqrt{25}.\sqrt{24}}\\ =1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\\ =1-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\)
=1-1/5=4/5
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\(A=\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+....+\dfrac{1}{\sqrt{2}+1}\)
\(A=\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+......+\sqrt{2}-1=\sqrt{25}-1=4\)
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Ta có:\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)(Nhân liên hợp)
Áp dụng vào bài toán,ta có:
\(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+.....+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}\)
\(=\frac{9}{10}\)
1/((k+1)√k+k√(k+1))
=((k+1)√k-k√(k+1))/((k+1)^2.k-k^2(k+1)
=((k+1)√k-k√(k+1))/k(k+1)(k+1-k)
=1/√k-1/√(k+1)
1/(2+√2)=1-1/√2
1/(3√2+2√3)=1/√2-1√3
......
1/(100√99+99√100)=1/√99-1/√100
=>1/(2+√2)+1/(3√2+2√3)+......+1/(100√99+99√100)
=1-1/√2+1/√2-1√3+...+1/√99-1/√100
=1-1/√100=1-1/10=9/10
Vẽ đồ thị hàm số:
1, y = 1/4x mũ 2
2, y = -1/4 x mũ 2
3, y = -2 x mũ 2
4, y = -1/2 x mũ 2
5, y = 3 x mũ 2
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A = ((20 + 1) . 20 : 2) . 2 = 420
B = (25 + 20) . 6 : 2 = 135
C = ( 33 + 26) . 8 : 2 = 236
D = (1 + 100) .100 : 2 = 5050
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Chị tham khảo ở đây ạ:
Câu hỏi của Vũ Thảo Vy - Toán lớp 9 - Học toán với OnlineMath
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\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+25\sqrt{24}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}=\frac{4}{5}\)
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\(U\left(n\right)=\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}\)
\(U\left(n\right)=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n.\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{n\left(n+1\right)\left(n+1-n\right)}\)
\(U\left(n\right)=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}\sqrt{n+1}\right)^2}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(S_{u\left(n\right)}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}< 1\)