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已知多项式x^2+ax+y-b与bx^2-3x+6y-3的和的值与字母x无关,求代数式3(a^2-2ab-b^2)-(3a^2-4ab-4b^2)的值
已知x^2-3x-6=0,求x^3-5x^2+2012的值
人气:101 ℃ 时间:2019-08-17 17:28:00
解答
(x^2+ax+y-b)+(bx^2-3x+6y-3)
=x^2+ax+y-b+bx^2-3x+6y-3
=(b+1)x^2 +(a-3)x+7y-b-3
因为值与字母x无关,所以 b+1=0 ,a-3=0 ,所以a=3,b=-1
3(a^2-2ab-b^2)-(3a^2-4ab-4b^2)
=3a^2-6ab-3b^2-3a^2+4ab+4b^2
=-2ab+b^2
将a=3,b=-1代入,原式=-2*3*(-1) +(-1)^2 =7
已知x^2-3x-6=0,求x^3-5x^2+2012的值
x^2-3x-6=0
x^2=3x+6
x^3-5x^2+2012
=x(x^2)-5x^2+2012
=x(3x+6)-5x^2 +2012
=3x^2 +6x-5x^2+2012
=-2x^2+6x+2012
=-2(3x+6)+6x+2012
=-6x-12+6x+2012
=2000对不起啊,麻烦帮忙算两道计算题:5ab-2[3ab-(4ab^2+2/1ab)]-5ab^26/5a^3b^2-3/1a^2b+5a^3b^2-3/1+6/5+3ab+3/2ba^2好吧第二题。。5ab-2[3ab-(4ab^2+2/1ab)]-5ab^2=5ab-2(3ab-4ab^2-1/2 ab)-5ab^2=5ab-6ab+8ab^2+ab-5ab^2=3ab^26/5a^3b^2-3/1a^2b+5a^3b^2-3/1+6/5+3ab+3/2ba^2(估计第一个数是6分之5,一般都写成5/6)=35/6 a^3b^2 - a^2b+3ab +1/2 (6分之35) (2分之1)
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