初二下册数学：计算题。先谢了。
\u521d\u4e8c\u4e0b\u518c\u6570\u5b66\uff1a\u8ba1\u7b97\u9898\u3002\u5148\u8c22\u4e86\u3002\uff082\uff09\uff08x²-4y²\uff09/(x²+4x+4) \u00d7 \uff08x+2\uff09/(3x²+6xy)
=\uff08x+2y\uff09\uff08x-2y\uff09/(x+2)² \u00d7 \uff08x+2\uff09/3x\uff08x+2y\uff09
=\uff08x-2y\uff09/(x+2) \u00d7 1/3x
=\uff08x-2y\uff09/(3x²+6x)
\uff084\uff09\uff08y²-x²\uff09/(5x²-4xy) \u00f7 \uff08x+y\uff09/\uff085x-4y\uff09
=\uff08y+x\uff09(y-x) / x(5x-4y) \u00d7 \uff085x-4y\uff09/\uff08x+y\uff09
=(y-x)/x
=5/2b^2
2.原式=-(a+9)(a-9)/(a+3)^2*2(a+3)/(a-9)*(a+3)/(a+9)
=-2
(1)(5ac)/(3bd)*(3d)/(2abc)=(15acd)/(6ab²cd)=(5/(2b²)
(2)=[(9-a)(9+a)(a+9)2(a+3)]/[(a+3)(a+3)(a-9)(a+3)]=-[(a+9)/(a+3)]²
（1）解：原式=(5ac/3bd)*(3d/2abc) （2）解：原式=[2（9+a)/a+3]*(a+3/a+9)
=5/2b² =2
（1）原式等于:4a²b/3cd²×5c²d/4ab²×3d/2abc=(4a²b×5c²d×3d)/(3cd²×4ab²×2abc)这时可以约掉分号上下相同的数字与字母，所以可得原式等于5/2b²
绛旓細1.鍘熷紡锛4a^2b/3cd^2*5c^2d/4ab^2*3d/2abc =5/2b^2 2.鍘熷紡=-(a+9)(a-9)/(a+3)^2*2(a+3)/(a-9)*(a+3)/(a+9)=-2
绛旓細=(2a²+ab+2b²)/(3a²-3b²)
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