先化简,在求值两道题
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先化简,在求值两道题
![](http://img.wesiedu.com/upload/0/52/0521dcf759d4c358b7a0b5252ac2dda2.jpg)
![](http://img.wesiedu.com/upload/0/52/0521dcf759d4c358b7a0b5252ac2dda2.jpg)
![先化简,在求值两道题](/uploads/image/z/17476595-35-5.jpg?t=%E5%85%88%E5%8C%96%E7%AE%80%2C%E5%9C%A8%E6%B1%82%E5%80%BC%E4%B8%A4%E9%81%93%E9%A2%98)
∵a=2
∴原式=[(a+1)(a-1)(a-3)(a+3)]/[(a+3)²(a+1)(a-1)]=(a-3)/(a+3)=(2-3)/(2+3)=-1/5
∵x=3
∴原式=[(x+2)(x-2)+(x-1)x][x(x-2)²]/[(x-2)²x(x-4)]=[x²-4+x²-x]/(x-4)=2x²-x-4)/(x-4)=2*3²-3-4)/(3-4)=-11
∴原式=[(a+1)(a-1)(a-3)(a+3)]/[(a+3)²(a+1)(a-1)]=(a-3)/(a+3)=(2-3)/(2+3)=-1/5
∵x=3
∴原式=[(x+2)(x-2)+(x-1)x][x(x-2)²]/[(x-2)²x(x-4)]=[x²-4+x²-x]/(x-4)=2x²-x-4)/(x-4)=2*3²-3-4)/(3-4)=-11