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  • 當(dāng)x=9/2時(shí),求(x^2+7x+10)/(x^2+6x+5)×(x^3+1)/(x^2+4x+4)÷(x^2-x+1)/(x+2)

    當(dāng)x=9/2時(shí),求(x^2+7x+10)/(x^2+6x+5)×(x^3+1)/(x^2+4x+4)÷(x^2-x+1)/(x+2)
    數(shù)學(xué)人氣:347 ℃時(shí)間:2020-03-24 21:53:06
    優(yōu)質(zhì)解答
    (x^2+7x+10)/(x^2+6x+5)×(x^3+1)/(x^2+4x+4)÷(x^2-x+1)/(x+2)
    =(x+2)(x+5)/(x+1)(x+5)*(x+1)(x^2-x+1)/(x+2)^2÷(x^2-x+1)/(x+2)
    =(x+2)(x+5)/(x+1)(x+5) * (x+1)(x^2-x+1)/(x+2)^2 * (x+2)/(x^2-x+1)
    =1(x+1)(x^2-x+1)/(x+2)^2是怎么來的?我不懂!謝謝了!x^3+1=(x+1)(x^2-x+1) 依據(jù):立方和公式,a^3+b^3=(a+b)(a^2-ab+b^2)x^2+4x+4=(x+2)^2即(x+2)² 依據(jù):完全平方公式,(a+b)^2=a^2+2ab+b^2逆用
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