早教吧作业答案频道 -->数学-->
化简求值[(x^2-16/x^2+8x+16)+(x/x-4)]/(1/x^2-16),其中x=根号2+1
题目详情
化简求值[(x^2-16/x^2+8x+16)+(x/x-4)]/(1/x^2-16),其中x=根号2+1
▼优质解答
答案和解析
原式=[(x+4)(x-4)/(x+4)²+x/(x-4)[*(x²-16)
=[(x-4)/(x+4)+x/(x-4)[*(x+4)(x-4)
=(x-4)/(x+4)*(x+4)(x-4)+x/(x-4)*(x+4)(x-4)
=(x-4)²+x(x+4)
=x²-8x+16+x²+4x
=2x²-4x+16
=2(3+2√2)-4(√2+1)+16
=18
=[(x-4)/(x+4)+x/(x-4)[*(x+4)(x-4)
=(x-4)/(x+4)*(x+4)(x-4)+x/(x-4)*(x+4)(x-4)
=(x-4)²+x(x+4)
=x²-8x+16+x²+4x
=2x²-4x+16
=2(3+2√2)-4(√2+1)+16
=18
看了化简求值[(x^2-16/x^...的网友还看了以下：