HOC24
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Chủ đề / Chương
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$\begin{array}{l} {x^3} + a{x^2} + bx + c = \left( {x + 1} \right)P\left( x \right) + 2021\\ \Rightarrow P\left( { - 1} \right) = 2021 \Rightarrow - 1 + a - b + c = 2021\\ {x^3} + a{x^2} + bx + c = \left( {x - 2} \right)P\left( x \right) + 2030\\ \Rightarrow P\left( 2 \right) = 2030 \Rightarrow 8 + 4a + 2b + c = 2030 \end{array}$
$\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} 4a + 2b + c = 2022\\ a - b + c = 2022 \end{array} \right. \Rightarrow 4a + 2b + c = a - b + c\\ \Rightarrow 3a + 3b = 0 \Leftrightarrow a = - b\\ \Rightarrow K = \left( {{a^{2021}} + {b^{2021}}} \right)\left( {{a^{2022}} + {b^{2022}}} \right) = \left( {{a^{2021}} - {a^{2021}}} \right)\left( {{a^{2022}} + {b^{2022}}} \right)\\ = 0\left( {{a^{2022}} + {b^{2022}}} \right) = 0 \end{array}$
b) Đặt $n^2-n+5=k^2(k\in \mathbb Z)$
$\begin{array}{l} \Rightarrow 4{n^2} - 4n + 20 = 4{k^2}\\ \Rightarrow {\left( {2n - 1} \right)^2} + 19 = {\left( {2k} \right)^2}\\ \Rightarrow \left( {2k - 2n + 1} \right)\left( {2k + 2n - 1} \right) = 19 \end{array}$
$\begin{array}{l} k \in \mathbb Z,n \in \mathbb Z \to 2k - 2n + 1,2k + 2n - 1 \in \mathbb Z\\ \bullet \left\{ \begin{array}{l} 2k - 2n + 1 = 1\\ 2k + 2n - 1 = 19 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 2k = 2n\\ 2n + 2n = 20 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} k = \dfrac{{20}}{3}\\ n = \dfrac{{10}}{3} \end{array} \right.\left( L \right) \end{array}$
$\begin{array}{l} \bullet \left\{ \begin{array}{l} 2k - 2n + 1 = - 1\\ 2k + 2n - 1 = - 19 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2k = 2n - 2\\ 2k + 2n = - 18 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} k = - 5\\ n = - 4 \end{array} \right.\left( {tm} \right)\\ \bullet \left\{ \begin{array}{l} 2k - 2n + 1 = 19\\ 2k + 2n - 1 = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} k = 5\\ n = - 4 \end{array} \right.\left( {tm} \right)\\ \bullet \left\{ \begin{array}{l} 2k - 2n + 1 = - 19\\ 2k + 2n - 1 = - 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} k = - 5\\ n = 5 \end{array} \right.\left( {tm} \right) \end{array}$
Vậy $n=-4, n=5$ thỏa mãn yêu cầu bài toán.
cos2x+√3sin2x+√3sinx−cosx=4cos2x+3sin2x+3sinx−cosx=4
Cho các số dương x, y, z thỏa mãn: xy+zx+4yz=32
Tìm giá trị nhỏ nhất của P=x^2+16y^2+16z^2
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