已知x=4-3,求x4-6x3-2x2+18x+23x2-8x+15的值
3个回答
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已扒庆知得(x-4)2=3,即x2-8x+13=0,则x2-8x=-13.
分子x4-6x3-2x2+18x+23,
=x4-8x3+2x3-2x2+18x+23,
=x2(x2-8x)+2x3-2x2+18x+23,
=-13x2+2x3-2x2+18x+23,
=2x3-16x2+x2+18x+23,
=2x(x2-8x)+x2+18x+23,
=-26x+x2+18x+23,
=x2-8x+23,档此数
=-13+23,
=10,行首
分母是x2-8x+15=-13+15=2,
∴
=
=5.
故答案为:5.
分子x4-6x3-2x2+18x+23,
=x4-8x3+2x3-2x2+18x+23,
=x2(x2-8x)+2x3-2x2+18x+23,
=-13x2+2x3-2x2+18x+23,
=2x3-16x2+x2+18x+23,
=2x(x2-8x)+x2+18x+23,
=-26x+x2+18x+23,
=x2-8x+23,档此数
=-13+23,
=10,行首
分母是x2-8x+15=-13+15=2,
∴
x4-6x3-2x2+18x+23 |
x2-8x+15 |
10 |
2 |
故答案为:5.
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