初二下册数学计算题.(1)计算:(-a/b)²÷(2a²/5b)²×a/5b;(2)计算:(x²+1)/(x²-1)-(x-2)/(x-1)÷(x-2)/x.(3)解方程:2/(x²-4)-1/(x+2)=0.

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初二下册数学计算题.(1)计算:(-a/b)²÷(2a²/5b)²×a/5b;(2)计算:(x²+1)/(x²-1)-(x-2)/(x-1)÷(x-2)/x.(3)解方程:2/(x²-4)-1/(x+2)=0.
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初二下册数学计算题.(1)计算:(-a/b)²÷(2a²/5b)²×a/5b;(2)计算:(x²+1)/(x²-1)-(x-2)/(x-1)÷(x-2)/x.(3)解方程:2/(x²-4)-1/(x+2)=0.
初二下册数学计算题.
(1)计算:(-a/b)²÷(2a²/5b)²×a/5b;
(2)计算:(x²+1)/(x²-1)-(x-2)/(x-1)÷(x-2)/x.
(3)解方程:2/(x²-4)-1/(x+2)=0.

初二下册数学计算题.(1)计算:(-a/b)²÷(2a²/5b)²×a/5b;(2)计算:(x²+1)/(x²-1)-(x-2)/(x-1)÷(x-2)/x.(3)解方程:2/(x²-4)-1/(x+2)=0.

(1)5/4ab
(2) -1/(1+x)
(3)x=4

我只想说 (2)应是-1/(x+1)

(1)5/4ab
(2)-1/x+1
(3)x=4

(1):(-a/b)²÷(2a²/5b)²×a/5b=5/4ab
(2):(x²+1)/(x²-1)-(x-2)/(x-1)÷(x-2)/x=-1/x+1
(3):2/(x²-4)-1/(x+2)=0 x=4

(1)(-a/b)²÷(2a²/5b)²×a/5b
=a2/b2÷4a4/25b2×a/5b
=a2/b2×25b2/4a4×a/5b
=5/4ab
(2)(x²+1)/(x²-1)-(x-2)/(x-1)÷(x-2)/x
=(x²+1)/(x²-1)-(x-2)/(x-1)×[x/(x-...

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(1)(-a/b)²÷(2a²/5b)²×a/5b
=a2/b2÷4a4/25b2×a/5b
=a2/b2×25b2/4a4×a/5b
=5/4ab
(2)(x²+1)/(x²-1)-(x-2)/(x-1)÷(x-2)/x
=(x²+1)/(x²-1)-(x-2)/(x-1)×[x/(x-2)]
=(x²+1)/(x²-1)-x/(x-1)
=(x²+1)/(x-1)(x+1)-x(x+1)/(x-1)(x+1)
=-1/(x+1)
(3)2/(x²-4)-1/(x+2)=0.
2/(x²-4)=1/(x+2)
方程两边同乘以(x²-4)得:
x-2=2
所以x=4
检验:x=4是原方程的解

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(1)
原式=a²/b²÷4a⁴/25b²×a/5b
=a²/b²×25b²/4a⁴×a/5b
=5/4ab
(2)
原式=(x²+1)/(x²-1)-(x-2)/(x-1)×[x/(x-2)]
=(x&#...

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(1)
原式=a²/b²÷4a⁴/25b²×a/5b
=a²/b²×25b²/4a⁴×a/5b
=5/4ab
(2)
原式=(x²+1)/(x²-1)-(x-2)/(x-1)×[x/(x-2)]
=(x²+1)/(x²-1)-x/(x-1)
=(x²+1)/(x-1)(x+1)-x(x+1)/(x-1)(x+1)
=-1/(x+1)
(3) 2/(x²-4)-1/(x+2)=0
2/(x+2)(x-2)-1/(x+2)=0
2-(x-2)=0
x=4
经检验得,x=4是原方程的解.

收起

(1)5/4ab
(2) 0
(3)x=4