x^2-8x+13=0求(x^4-16x^3+61x^2+24x+2)/(x 62-8x+15)

rr0072022-10-04 11:39:541条回答

x^2-8x+13=0求(x^4-16x^3+61x^2+24x+2)/(x 62-8x+15)
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馥藜共回答了25个问题 | 采纳率96%: x^2-8x+13=0
x^2-8x=-13
分子=x^4-8x^3-8x^3+61x^2+24x+2
=x^2(x^2-8x)-8x^3+61x^2+24x+2
=-13x^2-8x^3+61x^2+24x+2
=-8x^3+48x^2+24x+2
=-8x^3+64x^2-16x^2+24x+2
=-8x(x^2-8x)-16x^2+24x+2
=104x-16x^2+24x+2
=-16x^2+128x+2
=-16(x^2-8x)+2
=16*13+2
=210
分母=x^2-8x+15
=-13+15
=2
所以(x^4-16x^3+61x^2+24x+2)/(x^2-8x+15)
=210/2
=105; 1年前