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Áp dụng tính chất với mọi \(n\in N\) ta có \(\left[n+x\right]=n+\left[x\right]\)
Với \(k\in N\)
- Xét \(n=4k\):
\(\left[\frac{4k+2}{4}\right]+\left[\frac{4k+4}{4}\right]+\left[\frac{4k-1}{2}\right]=\left[k+\frac{1}{2}\right]+\left[k+1\right]+\left[2k-\frac{1}{2}\right]\)
\(=k+\left[\frac{1}{2}\right]+k+1+2k+\left[\frac{-1}{2}\right]=k+k+1+2k-1=4k=n\)
- Với \(n=4k+1\)
\(\left[\frac{4k+3}{4}\right]+\left[\frac{4k+5}{4}\right]+\left[\frac{4k}{2}\right]=\left[k+\frac{3}{4}\right]+\left[k+1+\frac{1}{4}\right]+\left[2k\right]\)
\(=k+\left[\frac{3}{4}\right]+k+1+\left[\frac{1}{4}\right]+2k=4k+1=n\)
- Với \(n=4k+2\)
\(\left[\frac{4k+4}{4}\right]+\left[\frac{4k+6}{4}\right]+\left[\frac{4k+1}{2}\right]=\left[k+1\right]+\left[k+1+\frac{1}{2}\right]+\left[2k+\frac{1}{2}\right]\)
\(=k+1+k+1+\left[\frac{1}{2}\right]+2k+\left[\frac{1}{2}\right]=4k+2=n\)
- Với \(n=4k+3\)
\(\left[\frac{4k+5}{4}\right]+\left[\frac{4k+7}{4}\right]+\left[\frac{4k+2}{2}\right]=\left[k+1+\frac{1}{4}\right]+\left[k+1+\frac{3}{4}\right]+\left[2k+1\right]\)
\(=k+1+k+1+2k+1=4k+3=n\)
Vậy \(\left[\frac{n+2}{4}\right]+\left[\frac{n+4}{4}\right]+\left[\frac{n-1}{2}\right]=n\)
//Cách chia trường hợp này hơi dài, k biết có cách nào tốt hơn ko nữa
\(\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2-2\left(\frac{1}{n}-\frac{1}{n\left(n+1\right)}-\frac{1}{n+1}\right)}\)
=1+1/n-1/n+1
chúc bn hoc tốt
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=1+\frac{1}{n}-\frac{1}{n+1}\)
\(\Rightarrow\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}=\frac{[\left(n+1\right)^2-n]^2}{n^2\left(n+1\right)^2}\)
\(\Rightarrow\left(n+1\right)^4+n^2=\left(n+1\right)^4-2\left(n+1\right)^2n+n^2\)
\(\Rightarrow0=-2\left(n+1\right)^2n\)
\(\Rightarrow\orbr{\begin{cases}\left(n+1\right)^2=0\\n=0\end{cases}}\Rightarrow\orbr{\begin{cases}n=-1\\n=0\end{cases}}\) mà \(n\inℕ^∗\)
=> n\(\in\varnothing\)
bác tính thi chuyên toán hả
uh, bạn giúp mình nhé