Tiếng Việt lớp 1 lạ nhỉ

\(\left(m+n+q\right)^2=m^2+n^2+q^2\)

<=>\(m^2+n^2+q^2+2\left(mn+nq+qm\right)=m^2+n^2+q^2\)

<=>\(mn+nq+qm=0\)

<=>\(\frac{mn+nq+qm}{mnq}=0\)

<=>\(\frac{mn}{mnq}+\frac{nq}{mnq}+\frac{qm}{mnq}=0\)

<=>\(\frac{1}{q}+\frac{1}{m}+\frac{1}{n}=0\)

<=>\(\frac{1}{m}+\frac{1}{n}=-\frac{1}{q}\)

<=>\(\left(\frac{1}{m}+\frac{1}{n}\right)^3=\left(-\frac{1}{q}\right)^3\)

<=>\(\frac{1}{m^3}+\frac{3}{mn}\left(\frac{1}{m}+\frac{1}{n}\right)+\frac{1}{n^3}=-\frac{1}{q^3}\)

<=>\(\frac{1}{m^3}+\frac{1}{n^3}+\frac{1}{q^3}=-\frac{3}{mn}\cdot\left(-\frac{1}{q}\right)=\frac{3}{mnq}\) (đpcm)

M = 0

\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)-\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\)

\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)(1-1)\)

\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right).0\)

\(M=0\)

Vì số bị trừ và số trừ gồm hai tích đảo ngược nhau nên M=0

1/ Trước hết ta chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)

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

Áp dụng : 

\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

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

Với mọi \(n\ge2\)

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

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

Lại có : \(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=\frac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n}+\sqrt{n-1}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\)

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

Từ (1) và (2) suy ra \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)

Áp dụng với n = 2,3,4,...,100 được đpcm.

mik đag cần gấp các bn giải nhanh dùm mik nha