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?Amanda?, Phạm Lan Hương, Phạm Thị Diệu Huyền, Vũ Minh Tuấn, Nguyễn Ngọc Lộc , @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @Trần Thanh Phương
giúp e với ạ! Cần trước 5h chiều nay! Cảm ơn mn nhiều!
Tranh thủ làm 1, 2 bài rồi ăn cơm:
1/ Đặt \(m=n-2008>0\)
\(\Rightarrow2^{2008}\left(369+2^m\right)\) là số chính phương
\(\Rightarrow369+2^m\) là số chính phương
m lẻ thì số trên chia 3 dư 2 nên ko là số chính phương
\(\Rightarrow m=2k\Rightarrow369=x^2-\left(2^k\right)^2=\left(x-2^k\right)\left(x+2^k\right)\)
b/
\(2\left(a^2+b^2\right)\left(a+b-2\right)=a^4+b^4\) \(\left(a+b>2\right)\)
\(\Rightarrow2\left(a^2+b^2\right)\left(a+b-2\right)\ge\frac{1}{2}\left(a^2+b^2\right)^2\)
\(\Rightarrow a^2+b^2\le4\left(a+b-2\right)\)
\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2\le0\Rightarrow a=b=2\)
\(\Rightarrow x=y=4\)
- Với \(x=0\Rightarrow144>0\) (đúng)
- Với \(x\ne0\)
\(VT=\left(x-2\right)\left(x-6\right)\left(x+3\right)\left(x+4\right)+57x^2\)
\(=\left(x^2+12-8x\right)\left(x^2+12+7x\right)+57x^2\)
\(=x^2\left[\left(x+\frac{12}{x}-8\right)\left(x+\frac{12}{x}+7\right)+57\right]\)
\(=x^2\left[\left(x+\frac{12}{x}-8\right)^2+15\left(x+\frac{12}{x}-8\right)+57\right]\)
\(=x^2\left[\left(x+\frac{12}{x}-8+\frac{15}{2}\right)^2+\frac{3}{4}\right]>0;\forall x\ne0\)
Vậy...
a: ĐKXĐ: |x+1|<>|x-1|
=>x+1<>1-x
=>2x<>0
hay x<>0
Vậy: D=R\{0}
b: \(f\left(-x\right)=\dfrac{\left|-x+1\right|+\left|-x-1\right|}{\left|-x+1\right|-\left|-x-1\right|}=\dfrac{\left|x-1\right|+\left|x+1\right|}{\left|x-1\right|-\left|x+1\right|}\)
\(=-\dfrac{\left|x-1\right|+\left|x+1\right|}{\left|x+1\right|-\left|x-1\right|}=-f\left(x\right)\)
Đặt \(x=\left[x\right]+\left\{x\right\}\)
\(\Rightarrow\left[3x\right]=\left[3\left[x\right]+3\left\{x\right\}\right]=3\left[x\right]+\left[3\left\{x\right\}\right]\)
\(\left[x+\frac{2}{3}\right]=\left[\left[x\right]+\left\{x\right\}+\frac{2}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{2}{3}\right]\)
\(\left[x+\frac{1}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
\(\Rightarrow\left[x+\frac{2}{3}\right]+\left[x+\frac{1}{3}\right]+\left[x\right]=3\left[x\right]+\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
Ta cần chứng minh \(\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
- Nếu \(\frac{2}{3}\le\left\{x\right\}< 1\Rightarrow\left\{{}\begin{matrix}2\le\left[3\left\{x\right\}\right]< 3\\1\le\left[\left\{x\right\}+\frac{2}{3}\right]< 2\\1\le\left[\left\{x\right\}+\frac{1}{3}\right]< 2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=2\\\left[\left\{x\right\}+\frac{2}{3}\right]=1\\\left[\left\{x\right\}+\frac{1}{3}\right]=1\end{matrix}\right.\)
\(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
- Nếu \(\frac{1}{3}\le\left\{x\right\}< \frac{2}{3}\Rightarrow\left\{{}\begin{matrix}1\le\left[3\left\{x\right\}\right]< 2\\1\le\left[\left\{x\right\}+\frac{2}{3}\right]< 2\\0\le\left[\left\{x\right\}+\frac{1}{3}\right]< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=1\\\left[\left\{x\right\}+\frac{2}{3}\right]=1\\\left[\left\{x\right\}+\frac{1}{3}\right]=0\end{matrix}\right.\)
\(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
- Nếu \(0< \left\{x\right\}< \frac{1}{3}\) tương tự trên ta có:
\(\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=0\\\left[\left\{x\right\}+\frac{2}{3}\right]=0\\\left[\left\{x\right\}+\frac{1}{3}\right]=0\end{matrix}\right.\) \(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
(x-1)(x-3)(x-4)(x-6)+10=(x-1)(x-6)(x-3)(x-4)+10
=(x2-7x+6)(x2-7x+12)+10 (*)
Đặt x2-7x+9=a
\(\Rightarrow\)(*)\(\Leftrightarrow\) (a-3)(a+3)+10=a2-9+10=a2+1\(\ge\)1 với mọi x