初2的4道数学计算题http://zhidao.baidu.com/question/227719887.html#sh

2个回答

  • (x-y)(1/9x-y)-(1/3-y)(1/3+y)

    =1/9x^2-xy-1/9xy+y^2-1/9+y^2

    =1/9x^2-10/9xy+2y^2-1/9

    =1/9(x^2-10xy+18y^2)

    (1+1/2)(1+1/2)(1+1/24)(1+1/28)

    =1/4(2+1)(2+1)(1+1/24)(1+1/28)

    =9/4(1+1/24)(1+1/28)

    =9/96(24+1)(1+1/28)

    =9/2688 (24+1)(28+1)

    =9×25×29÷2688

    =2.427

    已知√a=√b+1

    求(√a+√b)(a+b)(a^2+b^2)(a^4+b^4)

    ∵√a=√b+1

    ∴√a-√b=1

    a-b=(√a+√b)(√a-√b)=(√a+√b)

    (√a+√b)(a+b)(a^2+b^2)(a^4+b^4)

    =(a-b)(a+b)(a^2+b^2)(a^4+b^4)

    =(a^2-b^2)(a^2+b^2)(a^4+b^4)

    =(a^4-b^4)(a^4+b^4)

    =a^8-b^8

    已知a^2+b^2+4c^2-2a-4b+4c=-6

    求a-b+c的值

    a^2+b^2+4c^2-2a-4b+4c=-6

    a^2+b^2+4c^2-2a-4b+4c+6=0

    (a^2-2a+1)+(b^2-4b+4)+(4c^2+4c+1)=0

    (a-1)^2+(b-2)^2+(2c+1)^2=0

    则 a=1,b=2,c=-0.5

    把 a=1,b=2,c=-0.5 代进

    a-b+c 得:

    a-b+c=1-2-0.5=-1.5