(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...
(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)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
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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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
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
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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
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
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