![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt
P =1^2002 + 2^2002 + 3^2002 +4^2002 +...+ 2002^2002
Q = 1^2+2^2+..+ 2002^2, ta có Q = 1/6*2002*2003*(2.2002+1) ≡ 0 (mod 11)
{Công thức 1^2 +2^2 +...+ n^2 = n(n+1)(2n+1)/6}
P - Q = (1^2002 -1^2) + (2^2002-2^2) +..+ (2^2002 -2002^2)
Theo định lý Fermat nhỏ thì a^(p-1) ≡ 1 (mod p)
=> a^10 ≡ 1 (mod 11)
=> a^2000 ≡ 1 (mod 11)
=> a^2002 ≡ a^2 (mod 11) (*)
Từ (*) => P - Q ≡ 0 (mod 11)
mà Q ≡ 0 (mod 11) theo cm trên
=> P ≡ 0 (mod 11)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét với n là số tự nhiên không nhỏ hơn 1
Ta có : \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng điều trên ta có
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2002\sqrt{2001}+2001\sqrt{2002}}\)
\(=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2002}}\)
\(=1-\frac{1}{\sqrt{2002}}< 1-\frac{1}{\sqrt{2025}}=1-\frac{1}{45}=\frac{44}{45}\)
ta chứng minh công thức tổng quát sau
\(\frac{1}{\left[n+1\right]\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left[n+1\right]}\left[\sqrt{n+1}+\sqrt{n}\right]}\)
=\(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left[n+1\right]}\left[n+1-n\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left[n+1\right]}}\)
=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
ta có \(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
........
\(\frac{1}{2002\sqrt{2001}+2001\sqrt{2002}}=\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2002}}\)
=> \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+..+\frac{1}{2002\sqrt{2001}+2001\sqrt{2002}}\)
=\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2002}}\)
=\(1-\frac{1}{\sqrt{2002}}< \frac{44}{45}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{2002-1}{2002!}\right)+\frac{1}{2002!}\)
\(A=\left(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{2002}{2002!}-\frac{1}{2002!}\right)+\frac{1}{2002!}\)
\(A=\left(\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{2001!}-\frac{1}{2002!}\right)+\frac{1}{2002!}\)
\(A=\frac{1}{1!}-\frac{1}{2002!}+\frac{1}{2002!}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x^2-x+2001}=a\\\sqrt[3]{3x^2-7x+2002}=b\\\sqrt[3]{6x-2003}=c\end{matrix}\right.\)
\(\Rightarrow a^3-b^3-c^3=2002\) từ đây ta có:
\(a-b-c=\sqrt[3]{a^3-b^3-c^3}\)
\(\Leftrightarrow\left(a-b-c\right)^3=\sqrt[3]{a^3-b^3-c^3}\)
\(\Leftrightarrow\left(a-c\right)\left(a-b\right)\left(b+c\right)=0\)
Tự làm nốt nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt 2002=a; 2003=b
Theo đề, ta có:
\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}>\sqrt{a}+\sqrt{b}\)
\(\Leftrightarrow\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}>\sqrt{a}+\sqrt{b}\)
\(\Leftrightarrow a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a}>0\)
\(\Leftrightarrow a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)>0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(\sqrt{a}+\sqrt{b}\right)>0\)(luôn đúng)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2002}{\sqrt{2003}}+\frac{2003}{\sqrt{2002}}\)
=\(\frac{2002\sqrt{2003}}{\sqrt{2003}.\sqrt{2003}}+\frac{2003\sqrt{2002}}{\sqrt{2002}.\sqrt{2002}}\)
=\(\frac{\sqrt{2002}.\sqrt{2002}.\sqrt{2003}}{\sqrt{2003}.\sqrt{2003}}+\frac{\sqrt{2003}.\sqrt{2003}.\sqrt{2002}}{\sqrt{2002}.\sqrt{2002}}\)
>\(\frac{\sqrt{2002}.\sqrt{2002}.\sqrt{2003}+\sqrt{2003}.\sqrt{2003}.\sqrt{2002}}{\sqrt{2003}.\sqrt{2002}}\)
>\(\frac{\sqrt{2002}.\sqrt{2003}.\left(\sqrt{2002}+\sqrt{2003}\right)}{\sqrt{2003}.\sqrt{2002}}\)
>\(\sqrt{2002}+\sqrt{2003}\)
=>\(\frac{2002}{\sqrt{2003}}+\frac{2003}{\sqrt{2002}}\)>\(\sqrt{2002}+\sqrt{2003}\)(dpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\sqrt{x^2+2002}\)thì \(a^2=x^2+2002\Leftrightarrow a^2-x^2=2002\)
pt: \(\Leftrightarrow x^4+a=a^2-x^2\Leftrightarrow x^4-a^2+x^2+a=0\Leftrightarrow\left(x^2-a\right)\left(x^2+a\right)+\left(x^2+a\right)=0\)
\(\Leftrightarrow\left(x^2+a\right)\left(x^2-a+1\right)=0\)
\(x^2>0;a\ge\sqrt{2002}\)nên: \(x^2-a+1=0\Leftrightarrow x^2+1=\sqrt{x^2+2002}\)
Do 2 vế đều không âm nên ta bình phương 2 vế:\(x^4+2x^2+1=x^2+2002\Leftrightarrow x^4+x^2-2001=0\)
Tới đây pt trùng phương giải tiếp đi bn.