Question

七年级分式计算2道

1,已知x/y=3/4 ，求代数式的值

2,已知实数a满足a^2-5a+4=0,求代数式的值

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Answer 1

1.
原式
=(x+y)(x-y)/(x-y)² × x(x-y)/y(x+y)
=x/y
=3/4

2.
a²-5a+4=0
(a-1)(a-4)=0
a=1或a=4

原式
=4(a+3)/(3a-4)(a+2) ÷ [(a-2)- 5/(a+2)]
=4(a+3)/(3a-4)(a+2) × (a+2)/(a+3)(a-3)
=4/(3a-4)(a-3)
当a=1 ,原式=4/(3-4)(1-3)=2
当a=4 原式=4/(12-4)(4-3)=1/2

Answer 2

1，原式=（X+Y)(X-Y)/(X-Y)(X-Y) 除以Y(X+Y)/X(X-Y)约分后 =X/Y=3/4

2,原式=4（A+3)/(3A-4)(A+2) 除以[(a+3)(a-3)/(a+2)]=4/(3a-4)(a-3)

a^2-5a+4=0, 得 A=4或A=1, 所以原式=0.5 或2