Question

初二的数学计算题、写过程、谢谢亲故们

Answer 1

(6)原式=(b-a)/ab+(a^2-b^2)/ab
=[(b-a)+(a+b)(a-b)]/ab
=[(a+b-1)(a-b)]/ab
(7)原式=(x-1)/(x-1)^2+1/(x+1)(x-1)
=1/(x-1)+1/(x+1)(x-1)
=[(x+1)+1]/(x+1)(x-1)
=(x+2)/(x+1)(x-1)
将x=2^0.5-1代入
原式=(2^0.5+1)/[2^0.5×(2^0.5-2)]
=(根号2+1)/(2-二倍根号2)
(8)原式=(a-1-1)/(a-1)+(a-2)^2/(a^2-a)
=(a-2)/(a-1)+(a-2)^2/[a(a-1)]
=[a(a-2)+(a-2)^2)]/[a(a-1)]
将a=-1代入
原式=6
(9)原式化为3(x-1)=2x
3x-2x=3
x=3
(10)两边同时乘以x(x+1)得到
x^2+x^2-1=2x(x+1)
2x^2-1=2x^2+2x
2x=-1
x=-0.5

Answer 2

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