Toán Lớp 8: Tính nhân phân thức c)(x²+1)(x⁴+1)(x⁸+1)/x¹⁶-1 D) x²-6x+9/x²-3x+9. x³+27/3x-9

Question

Toán Lớp 8:  Tính nhân phân thức c)(x²+1)(x⁴+1)(x⁸+1)/x¹⁶-1 D) x²-6x+9/x²-3x+9. x³+27/3x-9

cobexinhdep · Answer

a,(x^2+1)(x^4+1)(x^8+1)/(x^16-1)     =(x^2+1)(x^4+1)(x^8+1)/(x^8+1)(x^8-1)     =(x^2+1)(x^4+1)/(x^4-1)(x^4+1)     =(x^2+1)/(x^2-1)(x^2+1)     =1/(x^2-1)        b,(x^2-6x+9)/(x^2-3x+9) . (x&sup3;+27)/(3x-9)     =((x-3)^2)/(x^2-3x+9) . ((x^3+3)(x^2-3x+9))/(3(x-3))     =((x-3)(x+3))/3=(x^2-9)/3

giangnguyen33 · Answer

Gửi em:   $c,$ $ dfrac{(x^2+1)(x^4+1)(x^8+1)}{x^{16}+1}$    $= dfrac{(x^2-1)(x^2+1)(x^4+1)(x^8+1)}{(x^2-1)(x^{16}+1)}$    $= dfrac{(x^4-1)(x^4+1)(x^8+1)}{(x^2-1)(x^{16}+1)}$    $= dfrac{(x^8-1)(x^8+1)}{(x^2-1)(x^{16}+1)}$    $= dfrac{(x^{16}-1)}{(x^2-1)(x^{16}+1)}$    $= dfrac{1}{x^2-1}$    $d,$ $ dfrac{x^2-6x+9}{x^2-3x+9}. dfrac{x^3+27}{3x-9}$    $= dfrac{(x-3)^2}{x^2-3x+9}. dfrac{(x+3)(x^2-3x+9)}{3(x-3)}$    $= dfrac{x-3}{x^2-3x+9}. dfrac{(x+3)(x^2-3x+9)}{3}$    $= dfrac{x-3}{1}. dfrac{x+3}{3}$    $= dfrac{x^2-9}{3}$