(1)已知a²-ab=2,4ab-3b³=-3,试求a²-13ab+9b²-5的值(2)已知m-n=2,mn=1试求多项式(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)的值
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![(1)已知a²-ab=2,4ab-3b³=-3,试求a²-13ab+9b²-5的值(2)已知m-n=2,mn=1试求多项式(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)的值](/uploads/image/z/3678437-29-7.jpg?t=%EF%BC%881%EF%BC%89%E5%B7%B2%E7%9F%A5a%26%23178%3B-ab%3D2%2C4ab-3b%26%23179%3B%3D-3%2C%E8%AF%95%E6%B1%82a%26%23178%3B-13ab%2B9b%26%23178%3B-5%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E5%B7%B2%E7%9F%A5m-n%3D2%2Cmn%3D1%E8%AF%95%E6%B1%82%E5%A4%9A%E9%A1%B9%E5%BC%8F%EF%BC%88-2mn%2B2m%2B3n%EF%BC%89-%EF%BC%883mn%2B2n-2m%EF%BC%89-%EF%BC%88m%2B4n%2Bmn%EF%BC%89%E7%9A%84%E5%80%BC)
(1)已知a²-ab=2,4ab-3b³=-3,试求a²-13ab+9b²-5的值(2)已知m-n=2,mn=1试求多项式(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)的值
(1)已知a²-ab=2,4ab-3b³=-3,试求a²-13ab+9b²-5的值
(2)已知m-n=2,mn=1试求多项式(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)的值
(1)已知a²-ab=2,4ab-3b³=-3,试求a²-13ab+9b²-5的值(2)已知m-n=2,mn=1试求多项式(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)的值
a²-ab=2 (1)
4ab-3b³=-3,(2)
(1)-(2)×3得:
a²-ab-12ab+9b²=2+9
a²-13ab+9b²-5=11-5=6
(2)已知m-n=2,mn=1
(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)
=-2mn+2m+3n-3mn-2n+2m-m-4n-mn
=3m-3n-6mn
=3(m-n)-6mn
=6-6
=0
a²-ab=2
12ab-9b²=-9
a²-ab-12ab+9b²=2+11
a²-13ab+9b²-5=13-5=8
(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)
=-6mn+3m-3n
=-6mn+3(m-n)
=-6+6
=0
6
(1)因为4ab-3b^2=-3
所以:12ab-9b^2=-9 (1)
因为a^2-ab=2 (2)
(2)-(1)
a^2-13ab+9b^2=11
a^2-13ab+9b^2-5=6
所以a^2-13ab+9b^2-5的值是6
(2)(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)
=-2mn+2m+3n-3...
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(1)因为4ab-3b^2=-3
所以:12ab-9b^2=-9 (1)
因为a^2-ab=2 (2)
(2)-(1)
a^2-13ab+9b^2=11
a^2-13ab+9b^2-5=6
所以a^2-13ab+9b^2-5的值是6
(2)(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)
=-2mn+2m+3n-3mn-2n+2m-m-4n-mn
=-6mn+3m-3n
=-6mn+3(m-n)
把m-n=2和mn=1代入-6mn+3(m-n)=-6+6=0
所以所求多项式的值是0
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