题目

因式分解 x=√2时 求X^2+7X+4/X^3-X^2-6X + X^2+6X+5/X^2-2X-3 的值

提问时间：2020-11-01

答案

(X²+7X+4)/(X³-X²-6X) + (X²+6X+5)/(X²-2X-3)
=(x²+7x+4)/x(x+2)(x-3)+(x+1)(x+5)/(x+1)(x-3)
=(x²+7x+4)/x(x+2)(x-3)+(x+5)/(x-3)
=(x²+7x+4+x³+7x²+10x)/x(x+2)(x-3)
=(x³+8x²+16x+x+4)/x(x+2)(x-3)
=[x(x+4)²+(x+4)]/x(x+2)(x-3)
=(x+4)(x²+4x+1)/x(x+2)(x-3)
=(√2+4)(2+4√2+1)/[√2(√2+2)(√2-3)]
=(2√2+8+8+16√2+√2+4)/(2√2+4-6-6√2)
=(19√2+20)/(-2-4√2)
=-(19√2+20)(2-4√2)/(4-32)
=-(19√2+20)(1-2√2)/(-14)
=(19√2+20-76-40√2)/(14)
=(-56-21√2)/14
=(-8-3√2)/2