Giải phương trình với a,b,c nguyên thỏa mãn:
$a(b-c)(b+c-a)^2+c(a-b)(a+b-c)^2=1$
Edited by Jinbe, 21-07-2013 - 10:38.
$a(b-c)(b+c-a)^2+c(a-b)(a+b-c)^2=1$
Started By kldlkvipmath, 20-07-2013 - 21:33
#2
Posted 21-07-2013 - 10:54
Zaraki
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PQT
Giải phương trình với a,b,c nguyên thỏa mãn:
$a(b-c)(b+c-a)^2+c(a-b)(a+b-c)^2=1 \qquad (1)$
Lời giải. Ta có $$\begin{aligned} (1) & \Leftrightarrow a(b-c)(b+c-a)^2+c(a-b)[(b+c-a)+2(a-c)]^2=1 \\ & \Leftrightarrow a(b-c)(b+c-a)^2+c(a-b)(b+c-a)^2+4c(a-b)(a-c)(b+c-a)+4c(a-b)(a-c)^2=1 \\ & \Leftrightarrow b(a-b)(b+c-a)^2+4c(a-b)(a-c)(b+c-a)+4c(a-b)(a-c)^2=1 \\ & \Leftrightarrow (a-c) \left[ b(b+c-a)^2+4c(a-b)(b+c-a)+4c(a-b)(a-c) \right] =1 \\ & \Leftrightarrow (a-c) \left \{ b \left[ (b+c-a)^2-4c(b+c-a)+4c^2 \right] -4c^2b+4ca(b+c-a)+4c(a-b)(a-c) \right \}=1 \\ & \Leftrightarrow (a-c)b(b-a-c)^2=1 \end{aligned}$$
Đến đây xét trường hợp là ra.
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