phân tích đa x64+x32+1 thành nhân tử
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\(\left(x^2-4\right)\left(x^2-10\right)-72\)
\(=\left(x^2-7+3\right)\left(x^2-7-3\right)-72\)
\(=\left(x^2-7\right)^2-9-72\)
\(=\left(x^2-7\right)^2-81\)
\(=\left(x^2-7+9\right)\left(x^2-7-9\right)\)
\(=\left(x^2+2\right)\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
\(\left(x^2-4\right)\left(x^2-10\right)-72\)
\(=x^4-10x^2-4x^2+40-72\)
\(=x^4-14x^2-32\)
\(=\left(x^2-16\right)\left(x^2+2\right)=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
\(x^4+6x^3+7x^2-6x+1\\ =\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)
`= x^2(x^2 + x + 1) - x(x^2 + x + 1) + 2024(x^2 + x + 1)`
`= (x^2 - x + 2024)(x^2 + x + 1)`.
a) TH1: x = 1
=> Giá tiền phải trả là: 11000 (đồng)
TH2: x > 1
=> Giá tiền phải trả là:
11000 + 10000(x - 1)
= 11000 + 10000x - 10000
= 10000x + 1000 (đồng) (1)
b) Người đó đi 50km ta thay x = 50 vào (1) ta có:
10000*50 + 1000
= 500000 + 1000
= 501000 (đồng)
\(x^{64}+x^{32}+1\\ =x^{64}+2x^{32}+1+x^{32}-2x^{32}\\ =\left[\left(x^{32}\right)^2+2\cdot x^{32}\cdot1+1^2\right]-x^{32}\\ =\left(x^{32}+1\right)^2-\left(x^{16}\right)^2\\ =\left(x^{32}-x^{16}+1\right)\left(x^{32}+x^{16}+1\right)\\ =\left(x^{32}-x^{16}+1\right)\left[\left(x^{32}+2x^{16}+1\right)+x^{16}-2x^{16}\right]\\ =\left(x^{32}-x^{16}+1\right)\left[\left(x^{16}+1\right)^2-x^{16}\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^{16}+x^8+1\right)\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left[\left(x^{16}+2x^8+1\right)-x^8\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left[\left(x^8+1\right)^2-x^8\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^8+x^4+1\right)\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left[\left(x^8+2x^4+1\right)-x^4\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left[\left(x^4+1\right)^2-x^4\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left[\left(x^4+2x^2+1\right)-x^2\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left[\left(x^2+1\right)^2-x^2\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(x^{64}+x^{32}+1\)
\(=x^{64}+2x^{32}-x^{32}+1\)
\(=\left(x^{64}+2^{32}+1\right)-x^{32}\)
\(=\left(x^{32}+1\right)^2-\left(x^{16}\right)^2\)
\(=\left(x^{32}+1-x^{16}\right)\left(x^{32}+1+x^{16}\right)\)