1、将一多项式[(17 x²-3x+4)(a x²+bx+c)],除以(5x+6)后,得商式为(2x+1),余式为0.求a-b-c=?2、已知(19x-31)(13x-17)-(13x-17)(11x-23)可因式分解成(ax+b)(8x+c),其中a、b、c均为整数,则a+b+c=?3、若|m-n|=n-m,且|m|=4
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![1、将一多项式[(17 x²-3x+4)(a x²+bx+c)],除以(5x+6)后,得商式为(2x+1),余式为0.求a-b-c=?2、已知(19x-31)(13x-17)-(13x-17)(11x-23)可因式分解成(ax+b)(8x+c),其中a、b、c均为整数,则a+b+c=?3、若|m-n|=n-m,且|m|=4](/uploads/image/z/9866983-31-3.jpg?t=1%E3%80%81%E5%B0%86%E4%B8%80%E5%A4%9A%E9%A1%B9%E5%BC%8F%5B%EF%BC%8817+x%26%23178%3B-3x%2B4%29%28a+x%26%23178%3B%2Bbx%2Bc%29%5D%2C%E9%99%A4%E4%BB%A5%285x%2B6%29%E5%90%8E%2C%E5%BE%97%E5%95%86%E5%BC%8F%E4%B8%BA%282x%2B1%29%2C%E4%BD%99%E5%BC%8F%E4%B8%BA0.%E6%B1%82a-b-c%3D%3F2%E3%80%81%E5%B7%B2%E7%9F%A5%2819x-31%29%2813x-17%29-%2813x-17%29%2811x-23%29%E5%8F%AF%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%E6%88%90%28ax%2Bb%29%288x%2Bc%29%2C%E5%85%B6%E4%B8%ADa%E3%80%81b%E3%80%81c%E5%9D%87%E4%B8%BA%E6%95%B4%E6%95%B0%2C%E5%88%99a%2Bb%2Bc%3D%3F3%E3%80%81%E8%8B%A5%7Cm-n%7C%3Dn-m%2C%E4%B8%94%7Cm%7C%3D4)
1、将一多项式[(17 x²-3x+4)(a x²+bx+c)],除以(5x+6)后,得商式为(2x+1),余式为0.求a-b-c=?2、已知(19x-31)(13x-17)-(13x-17)(11x-23)可因式分解成(ax+b)(8x+c),其中a、b、c均为整数,则a+b+c=?3、若|m-n|=n-m,且|m|=4
1、将一多项式[(17 x²-3x+4)(a x²+bx+c)],除以(5x+6)后,得商式为(2x+1),余式为0.求a-b-c=?
2、已知(19x-31)(13x-17)-(13x-17)(11x-23)可因式分解成(ax+b)(8x+c),其中a、b、c均为整数,则a+b+c=?
3、若|m-n|=n-m,且|m|=4,|n|=3则(m+n)²=?
4、若a=2007 / 2008,b=2008 / 2009试不用将分数化小数的方法比较a、b大小.
5、已知x²-5x=14,求(x-1)(2x-1)-(x+1)²+1的值
6、√6 ÷√3 =
1、将一多项式[(17 x²-3x+4)(a x²+bx+c)],除以(5x+6)后,得商式为(2x+1),余式为0.求a-b-c=?2、已知(19x-31)(13x-17)-(13x-17)(11x-23)可因式分解成(ax+b)(8x+c),其中a、b、c均为整数,则a+b+c=?3、若|m-n|=n-m,且|m|=4
2、已知(19x-31)(13x-17)-(13x-17)(11x-23)可因式分解成(ax+b)(8x+c),其中a、b、c均为整数,则a+b+c=?
(19x-31)(13x-17)-(13x-17)(11x-23)
=(13x-17)(19x-31-11x+23)
=(13x-17)(8x-8)
=(ax+b)(8x+c)
a=13,b=-17,c=8
3、若|m-n|=n-m,且|m|=4,|n|=3则(m+n)²=?
|m-n|=n-m,n>m
|m|=4,m=±4
|n|=3,n=±3
所以,n = ±3,m = -4
(m+n)²=(-4+3)²=1
或者,(m+n)²=(-4-3)²=49
4、若a=2007 / 2008,b=2008 / 2009试不用将分数化小数的方法比较a、b大小.
a = 2007 / 2008 = (2007 X 2009)/(2008 X 2009)=(2008² - 1)/(2008 X 2009)
b = 2008 / 2009 = (2008 X 2008 )/ (2008 X 2009)
所以,a<b
5、已知x²-5x=14,求(x-1)(2x-1)-(x+1)²+1的值
(x-1)(2x-1)-(x+1)²+1
= 2x² - 3x + 1 - x² -2x -1 + 1
= x² - 5x + 1
= 14 + 1
= 15
6、√6 ÷√3 = √2