1 :已知x²-5x=14 求(x-1)(2x-1)-(x+1)²+1的值
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  • 1) (x-1)(2x-1)-(x+1)2+1

    =2x2-3x+1-x2-2x-1+1

    =x2-5x+1

    =14+1

    =15

    2)C=2(2a+S/2a)=2(2a+2a-3b+1)=8a-6b+2

    3)(x+y)2=x2+2xy+y2=25,(x-y)2=x2-2xy+y2=9

    (x+y)2-(x-y)2=4xy=16

    xy=4

    (x+y)2+(x-y)2=2x2+2y2=34

    x2+y2=17

    4)a2+2b2+c2-2ab-2bc

    =a2-2ab+b2+c2-2bc+b2

    =(a-b)2+(c-b)2=0

    (a-b)2=0 (c-b)2=0

    a=b c=b

    a=b=c

    三角形为等边三角形

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