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Ta có 4 x n + 2 – 8 x n = 4 x n . x 2 – 8 x n = x n ( 4 x 2 – 8 )
Vậy khi đặt nhân tử chung x n ra ngoài ta được biểu thức còn lại là 4 x 2 – 8
Đáp án cần chọn là: B
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=(a+b+c)(a2+b2+c2−ab−bc−ca)
=(a+b+c)(a2+2ab+b2−ab−ac+c2)−3ab(a+b+c)
=(a+b)3+c3−3ab(a+b+c)
=a3+3ab(a+b)+b3+c3−3abc−3ab(a+b
a3+b3+c3−3abc
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\(M=32^{2023}-32^{2021}=32^{2021}\left(32^2-1\right)=32^{2021}.1023=32^{2021}.31.33\)
Vì \(31⋮31=>M⋮31\)
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\(a^{32}-b^{32}\)
\(=\left(a^{16}-b^{16}\right)\left(a^{16}+b^{16}\right)\)
\(=\left(a^8-b^8\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(=\left(a^4-b^4\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(a^{32}-b^{32}=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(4x^n+12.2x^n=32\)
\(4x^n+24x^n=32\)
\(28x^n=32\)
\(x^n=\frac{8}{7}\)
\(x=\sqrt[n]{\frac{8}{7}}\)