đề là tính tổng à

Viết ra thành \(M=1^2+2^2+3^2+...+99^2\) cho rồi chứ viết dấu * khó nhì quá;(

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Đặt \(A=\sqrt[3]{99-70\sqrt{2}}+\sqrt[3]{99+70\sqrt{2}}\)

Ta có: \(A^3=\left(\sqrt[3]{99-70\sqrt{2}}+\sqrt[3]{99+70\sqrt{2}}\right)^3\)

\(=99-70\sqrt{2}+99+70\sqrt{2}+3\cdot\sqrt[3]{\left(99-70\sqrt{2}\right)\left(99+70\sqrt{2}\right)}\cdot A\)

\(=198+3A\)

\(\Leftrightarrow A^3-198-3A=0\)

\(\Leftrightarrow A^3-3A-198=0\)

\(\Leftrightarrow A^3-6A^2+6A^2-36A+33A-198=0\)

\(\Leftrightarrow A^2\left(A-6\right)+6A\left(A-6\right)+33\left(A-6\right)=0\)

\(\Leftrightarrow\left(A-6\right)\left(A^2+6A+33\right)=0\)

mà \(A^2+6A+33>0\forall A\)

nên A-6=0

hay A=6

Vậy: \(\sqrt[3]{99-70\sqrt{2}}+\sqrt[3]{99+70\sqrt{2}}=6\)

Ta có

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}\)(nhân lượng liên hiệp nhé)

\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào bài toán ta có

\(\frac{1}{2\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)

Ta có:

\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\dfrac{99+1}{1\cdot99}+\dfrac{97+3}{3\cdot97}+...+\dfrac{1+99}{99\cdot1}}{100}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{99}+1\right)}{100}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{100}}=\dfrac{1}{\dfrac{2}{100}}=\dfrac{100}{2}=50\)

\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+...+\dfrac{1}{99}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{1+\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{100}{100}+\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)

Ta có \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

                                                                \(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)

                                                                \(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào A ta được

\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

    \(=1-\frac{1}{10}\)

   \(=\frac{9}{10}\)

Incursion_03 đúng mẹ nó rồi nhé!

tui cx định tl nhưng nó tl trước ns chung nó đúng cmnr

\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^2}+...+\frac{100}{2^{100}}\)

\(2A=2\left(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{100}{2^{100}}\right)=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{n-1}{2^n}+...+\frac{100}{2^{99}}\)

\(2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)

Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2B=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(\Rightarrow2B-B=B=2-\frac{1}{2^{99}}\)

\(\Rightarrow A=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}< 2\)

Chứng minh phụ: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\) (trục căn thức ở mẫu)

                                   \(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n^2+2n+1-n^2-n\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào tính: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(\frac{1}{\left(1+1\right)\sqrt{1}+1\sqrt{1+1}}+\frac{1}{\left(1+2\right)\sqrt{2}+2\sqrt{2+1}}+...+\frac{1}{\left(99+1\right)\sqrt{99}+99\sqrt{99+1}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

= 1 - 1/ căn 100

=1 - 1/10

= 9/10

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