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\(f\left(x\right)=ax^2+bx+2020\\ \Leftrightarrow f\left(\sqrt{3}-1\right)=a\left(4-2\sqrt{3}\right)+b\left(\sqrt{3}-1\right)+2020=2021\\ \Leftrightarrow4a-2a\sqrt{3}+b\sqrt{3}-b-1=0\\ \Leftrightarrow\left(4a-b-1\right)-\sqrt{3}\left(2a-b\right)=0\\ \Leftrightarrow4a-b-1=\sqrt{3}\left(2a-b\right)\)
Vì a,b hữu tỉ nên \(4a-b-1;2a-b\) hữu tỉ
Mà \(\sqrt{3}\) vô tỉ nên \(\sqrt{3}\left(2a-b\right)\) hữu tỉ khi \(2a-b=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}4a-b-1=0\\2a-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\)
\(\Leftrightarrow f\left(1+\sqrt{3}\right)=\dfrac{1}{2}\left(4+2\sqrt{3}\right)+1+\sqrt{3}+2020=2023+2\sqrt{3}\)
\(f\left(x\right)=6x^3-7x^2-16x+m\)
Do \(f\left(x\right)\) chia hết \(2x-5\), theo định lý Bezout:
\(f\left(\dfrac{5}{2}\right)=0\Rightarrow6.\left(\dfrac{5}{2}\right)^3-7.\left(\dfrac{5}{2}\right)^2-16.\left(\dfrac{5}{2}\right)+m=0\)
\(\Rightarrow m=-10\)
Khi đó \(f\left(x\right)=6x^3-7x^2-16x-10\)
Số dư phép chia cho \(3x-2\):
\(f\left(\dfrac{2}{3}\right)=6.\left(\dfrac{2}{3}\right)^3-7.\left(\dfrac{2}{3}\right)^2-16.\left(\dfrac{2}{3}\right)-10=-22\)
Do chia hết , theo định lý Bezout:
Khi đó
Số dư phép chia cho :
\(f\left(2k-1\right)=\left[\left(2k-1\right)^2+2k-1+1\right]^2+1\)
\(=\left(4k^2+1-2k\right)^2+1=\left(4k^2+1\right)^2-4k\left(4k^2+1\right)+4k^2+1\)
\(=\left(4k^2+1\right)\left(4k^2-4k+2\right)=\left(4k^2+1\right)\left[\left(2k-1\right)^2+1\right]\)
\(f\left(2k\right)=\left(4k^2+1+2k\right)^2+1=\left(4k^2+1\right)^2+4k\left(4k^2+1\right)+4k^2+1\)
\(=\left(4k^2+1\right)\left(4k^2+4k+2\right)=\left(4k^2+1\right)\left[\left(2k+1\right)^2+1\right]\)
\(\Rightarrow\frac{f\left(2k-1\right)}{f\left(2k\right)}=\frac{\left(4k^2+1\right)\left[\left(2k-1\right)^2+1\right]}{\left(4k^2+1\right)\left[\left(2k+1\right)^2+1\right]}=\frac{\left(2k-1\right)^2+1}{\left(2k+1\right)^2+1}\)
\(\Rightarrow\frac{f\left(1\right).f\left(3\right).f\left(5\right)...f\left(2k-1\right)}{f\left(2\right).f\left(4\right).f\left(6\right)...f\left(2k\right)}=\frac{2}{10}.\frac{10}{16}.\frac{16}{50}...\frac{\left(2k-3\right)^2+1}{\left(2k-1\right)^2+1}.\frac{\left(2k-1\right)^2+1}{\left(2k+1\right)^2+1}=\frac{2}{\left(2k+1\right)^2+1}\)
\(\Rightarrow\frac{f\left(1\right)f\left(3\right)...f\left(2017\right)}{f\left(2\right)f\left(4\right)...f\left(2018\right)}=\frac{2}{2019^2+1}=\frac{1}{2038181}\)
\(f\left(x\right):\left(x-a\right)\) dư r1
\(\Leftrightarrow f\left(x\right)=\left(x-a\right)\cdot a\left(x\right)+r_1\\ \Leftrightarrow f\left(a\right)=r_1\)
Vì \(\left(x-a\right)\left(x-b\right)\) là đa thức bậc 2 nên có dư bậc 1
Gọi dư của \(f\left(x\right):\left(x-a\right)\left(x-b\right)\) là \(cx+d\)
\(\Leftrightarrow f\left(x\right)=\left(x-a\right)\left(x-b\right)\cdot c\left(x\right)+cx+d\\ \Leftrightarrow f\left(a\right)=ac+d=r_1\left(1\right)\\ f\left(x\right)=\left(x-a\right)\left(x-b\right)\cdot c\left(x\right)+cx+d\\ =\left(x-a\right)\left(x-b\right)\cdot c\left(x\right)+c\left(x-b\right)+bc+d\\ =\left(x-b\right)\left[\left(x-a\right)\cdot c\left(x\right)+c\right]+bc+d\)
Vì \(f\left(x\right):\left(x-b\right)\) dư r2 nên \(bc+d=r_2\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}bc+d=r_2\\ac+d=r_1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c\left(a-b\right)=r_1-r_2\\ac+d=r_1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c=\dfrac{r_1-r_2}{a-b}\\d=r_1-\dfrac{a\left(r_1-r_2\right)}{a-b}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=\dfrac{r_1-r_2}{a-b}\\d=\dfrac{ar_2-br_1}{a-b}\end{matrix}\right.\)
Vậy đa thức dư là \(\dfrac{r_1-r_2}{a-b}x+\dfrac{ar_2-br_1}{a-b}\)