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TS VÒNG 2 CHUYÊN TOÁN 18-19


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#1
toanhoc2017

toanhoc2017

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Tìm nghiệm nguyên của phương trình $x^3+(x+1)^3+...+(x+7)^3=y^3$



#2
Tea Coffee

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Ta có:

$x^{3}+(x+1)^{3}+...+(x+7)^{3}=y^{3}<=>\left [ x^{3} -1\right ]+\left [ (x+1)^{3}-1 \right ]+...+\left [ (x+7)^{3}-1 \right ]=y^{3}-8=(y^{3}-1)-7$

Nhận thấy $a\epsilon \mathbb{Z}=>a^{3}-1=a(a-1)(a+1)\vdots 6=>VT\vdots 6$

Nhưng $VP$ không chia hết cho $6$.

Phương trình vô nghiệm.


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Today is a gift. That’s why it’s called the present.





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