Chủ đề này có 1 trả lời

[Beautifulsunrise]

GPT nghiệm nguyên dương: $\left \lfloor \sqrt[3]{1} \right \rfloor+\left \lfloor \sqrt[3]{2} \right \rfloor+\left \lfloor \sqrt[3]{3} \right \rfloor+...+\left \lfloor \sqrt[3]{x^3+2x+4} \right \rfloor=7225$

[nguyenta98]

GPT nghiệm nguyên dương: $\left \lfloor \sqrt[3]{1} \right \rfloor+\left \lfloor \sqrt[3]{2} \right \rfloor+\left \lfloor \sqrt[3]{3} \right \rfloor+...+\left \lfloor \sqrt[3]{x^3+2x+4} \right \rfloor=7225$

Giải như sau:
Nhận xét $\left\lfloor\sqrt[3]{x^3}\right\rfloor+...+\left\lfloor\sqrt[3]{x^3+3x^2+3x}\right\rfloor=x+x+...+x(\text{3x^2+3x+1 số})=x(3x^2+3x+1)$
Do đó
$\left\lfloor\sqrt[3]{1}\right\rfloor+...+\left\lfloor\sqrt[3]{7}\right\rfloor=1(3.1^2+3.1+1)$
$\left\lfloor\sqrt[3]{8}\right\rfloor+...+\left\lfloor\sqrt[3]{26}\right\rfloor=2(3.2^2+3.1+1)$
$.....$
$\left\lfloor\sqrt[3]{(x-1)^3}\right\rfloor+...+\left\lfloor\sqrt[3]{x^3-1}\right\rfloor=(x-1)(3.(x-1)^2+3(x-1)+1)$
$\left\lfloor\sqrt[3]{x^3}\right\rfloor+...+\left\lfloor\sqrt[3]{x^3+2x+4}\right\rfloor=x(2x+5)$
Suy ra $1(3.1^2+3.1+1)+2(3.2^2+3.2+1)+...+(x-1)(3(x-1)^2+3(x-1)+1)+x(2x+5)=7225$
$\Rightarrow 3(1^3+2^3+...+(x-1)^3)+3(1^2+2^2+...+(x-1)^2)+(x-1)+x(2x+5)=7225$
$\Rightarrow 3\left[\dfrac{x(x-1)}{2}\right]^2+3.\dfrac{(x-1)x(2x-1)}{6}+x-1+x(2x+5)=7225$
Đến đây chắc đã ngon lành

Bài viết đã được chỉnh sửa nội dung bởi nguyenta98: 09-08-2012 - 21:45