y^3-x^3=91 tìm nghiệm nguyên của phương trình 30/09/2021 Bởi Sadie y^3-x^3=91 tìm nghiệm nguyên của phương trình
Đáp án: $(x,y)=\{(5,6);(-6,-5);(-3,4)(-4,3)\}$ Lời giải: $\eqalign{ & {y^3} – {x^3} = 91 \cr & \Leftrightarrow \left( {y – x} \right)\left( {{y^2} + yx + {x^2}} \right) = 91 \cr & \Rightarrow y – x\,\,\, \in \text{Ư}\left( {91} \right) = \left\{ { \pm 1; \pm 7; \pm 13; \pm 91} \right\} \cr & \text{Do }{y^2} + yx + {x^2} > 0 \Rightarrow y – x > 0 \cr & \text{TH1: }\left\{ \matrix{ y – x = 1 \hfill \cr {y^2} + yx + {x^2} = 91 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 1 \hfill \cr {\left( {x + 1} \right)^2} + \left( {x + 1} \right)x + {x^2} = 91 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 1 \hfill \cr 3{x^2} + 3x – 90 = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ y = x + 1 \hfill \cr \left[ \matrix{ x = 5 \hfill \cr x = – 6 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 5;\,\,y = 6 \hfill \cr x =- 6;\,\,y = – 5 \hfill \cr} \right.\,\,\left( {\text{thỏa mãn}} \right) \cr & \text{TH2: }\left\{ \matrix{ y – x = 7 \hfill \cr {y^2} + yx + {x^2} = 13 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 7 \hfill \cr {\left( {x + 7} \right)^2} + \left( {x + 7} \right)x + {x^2} = 13 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 7 \hfill \cr {x^2} + 14x + 49 + {x^2} + 7x + {x^2} = 13 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 7 \hfill \cr 3{x^2} + 21x + 36 = 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 7 \hfill \cr \left[ \matrix{ x = – 3 \hfill \cr x = – 4 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = – 3;\,\,y = 4 \hfill \cr x = – 4;\,\,y = 3 \hfill \cr} \right.\,\,\left( {\text{thỏa mãn}} \right) \cr & \text{TH3:}\left\{ \matrix{ y – x = 13 \hfill \cr {y^2} + yx + {x^2} = 7 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 13 \hfill \cr {\left( {x + 13} \right)^2} + \left( {x + 13} \right)x + {x^2} = 7 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 13 \hfill \cr {x^2} + 26x + 169 + {x^2} + 13x + {x^2} = 7 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 13 \hfill \cr 3{x^2} + 39x + 162 = 0\,\,\left( {\text{Vô nghiệm}} \right) \hfill \cr} \right. \cr & \text{TH4: }\left\{ \matrix{ y – x = 91 \hfill \cr {y^2} + yx + {x^2} = 1 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 91 \hfill \cr {\left( {x + 91} \right)^2} + \left( {x + 91} \right)x + {x^2} = 1 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 91 \hfill \cr {x^2} + 182x + 8281 + {x^2} + 91x + {x^2} = 1 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 91 \hfill \cr 3{x^2} + 182x + 8280 = 0\,\,\left( {\text{vô nghiệm}} \right) \hfill \cr} \right. \cr} $ Bình luận
Đáp án:
$(x,y)=\{(5,6);(-6,-5);(-3,4)(-4,3)\}$
Lời giải:
$\eqalign{ & {y^3} – {x^3} = 91 \cr & \Leftrightarrow \left( {y – x} \right)\left( {{y^2} + yx + {x^2}} \right) = 91 \cr & \Rightarrow y – x\,\,\, \in \text{Ư}\left( {91} \right) = \left\{ { \pm 1; \pm 7; \pm 13; \pm 91} \right\} \cr & \text{Do }{y^2} + yx + {x^2} > 0 \Rightarrow y – x > 0 \cr & \text{TH1: }\left\{ \matrix{ y – x = 1 \hfill \cr {y^2} + yx + {x^2} = 91 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 1 \hfill \cr {\left( {x + 1} \right)^2} + \left( {x + 1} \right)x + {x^2} = 91 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 1 \hfill \cr 3{x^2} + 3x – 90 = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ y = x + 1 \hfill \cr \left[ \matrix{ x = 5 \hfill \cr x = – 6 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 5;\,\,y = 6 \hfill \cr x =- 6;\,\,y = – 5 \hfill \cr} \right.\,\,\left( {\text{thỏa mãn}} \right) \cr & \text{TH2: }\left\{ \matrix{ y – x = 7 \hfill \cr {y^2} + yx + {x^2} = 13 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 7 \hfill \cr {\left( {x + 7} \right)^2} + \left( {x + 7} \right)x + {x^2} = 13 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 7 \hfill \cr {x^2} + 14x + 49 + {x^2} + 7x + {x^2} = 13 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 7 \hfill \cr 3{x^2} + 21x + 36 = 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 7 \hfill \cr \left[ \matrix{ x = – 3 \hfill \cr x = – 4 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = – 3;\,\,y = 4 \hfill \cr x = – 4;\,\,y = 3 \hfill \cr} \right.\,\,\left( {\text{thỏa mãn}} \right) \cr & \text{TH3:}\left\{ \matrix{ y – x = 13 \hfill \cr {y^2} + yx + {x^2} = 7 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 13 \hfill \cr {\left( {x + 13} \right)^2} + \left( {x + 13} \right)x + {x^2} = 7 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 13 \hfill \cr {x^2} + 26x + 169 + {x^2} + 13x + {x^2} = 7 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 13 \hfill \cr 3{x^2} + 39x + 162 = 0\,\,\left( {\text{Vô nghiệm}} \right) \hfill \cr} \right. \cr & \text{TH4: }\left\{ \matrix{ y – x = 91 \hfill \cr {y^2} + yx + {x^2} = 1 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 91 \hfill \cr {\left( {x + 91} \right)^2} + \left( {x + 91} \right)x + {x^2} = 1 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 91 \hfill \cr {x^2} + 182x + 8281 + {x^2} + 91x + {x^2} = 1 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ y = x + 91 \hfill \cr 3{x^2} + 182x + 8280 = 0\,\,\left( {\text{vô nghiệm}} \right) \hfill \cr} \right. \cr} $