初二分式数学题
\u521d\u4e8c\u5206\u5f0f\u6570\u5b66\u98981.
100/(2x) - 100/x = 4
2.
\u8bbe\u5546\u573a\u7b2c\u4e00\u6279\u886c\u886b\u8d2d\u8fdb\u91cf\u4e3ax\u4e07\u4ef6\uff0c\u7531\u9898\u610f\u5f97
8/x = 17.6/(2x) - 4
\u89e3\u5f97 x = 0.2
\u6240\u4ee5\u5546\u573a\u7b2c\u4e00\u6279\u886c\u886b\u8d2d\u8fdb\u91cf\u4e3a2000\u4ef6\uff0c\u7b2c\u4e8c\u6279\u886c\u886b\u8d2d\u8fdb\u91cf\u4e3a4000\u4ef6\uff0c\u4e00\u5171\u9500\u552e\u989d\u4e3a(2000+4000-150) * 58 + 150 * 58 * 0.8 = 346260 \u5143\u3002
\u6240\u4ee5\u5728\u8fd9\u4e24\u7b14\u751f\u610f\u4e2d\uff0c\u5546\u573a\u5171\u76c8\u5229 346260 - 80000 - 176000 = 90260 \u5143\u3002
\u7b2c\u4e00\u9898(m+5)/(m-1)\uff1e0\uff0c
\u2234m\uff1c-5\uff0c\u6216m\uff1e1
(m+5)/(m-1) = (m-1+6)/(m-1)=1+6/(m-1)\u4e3a\u6574\u6570
m-1=\u00b11\uff0c\u00b12\uff0c\u00b13\uff0c\u00b16
m=0\uff0c2\uff1b-1\uff0c3\uff1b-2\uff0c4\uff1b-5\uff0c7
\u53c8\uff1am\uff1c-5\uff0c\u6216m\uff1e1
\u2234m = 2\uff0c3\uff0c4\uff0c7
\u7b2c\u4e8c\u9898
\u2235-5\uff1cx\uff1c3
\u2234x+5\uff1e0 x-3\uff1c0
\u2234x+5\u7684\u7edd\u5bf9\u503c\u7b49\u4e8ex+5
x-3\u7684\u7edd\u5bf9\u503c\u7b49\u4e8e3-x
\u2234x+5\u5206\u4e4bx+5\u7684\u7edd\u5bf9\u503c\u7b49\u4e8e1
3-x\u5206\u4e4bx-3\u7684\u7edd\u5bf9\u503c\u7b49\u4e8e1
\u5f53x\u5c0f\u4e8e0\u65f6\uff0cx\u5206\u4e4bx\u7684\u7edd\u5bf9\u503c\u7b49\u4e8e\uff0d1
\u5f53x\u5927\u4e8e0\u65f6\uff0cx\u5206\u4e4bx\u7684\u7edd\u5bf9\u503c\u7b49\u4e8e1
\u2234x+5\u5206\u4e4bx+5\u7684\u7edd\u5bf9\u503c\u52a03-x\u5206\u4e4bx-3\u7684\u7edd\u5bf9\u503c\u52a0x\u5206\u4e4bx\u7684\u7edd\u5bf9\u503c\u7684\u503c\u4e3a1\u62163
所以由(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)得xy+yz+zx=-6
又z=2-x-y
2z=4-2x-2y
所以1/(xy+2z)=1/(xy+4-2x-2y)=1/[(xy-2x)-(2y-4)]=1/[x(y-2)-2(y-2)]=1/[(x-2)(y-2)]
同理1/(yz+2x)=1/[(y-2)(z-2)]，1/(xz+2y)=1/[(x-2)(z-2)]
所以1/(xy+2z)+1/(yz+2x)+1/(xz+2y)
=1/[(x-2)(y-2)]+1/[(y-2)(z-2)]+1/[(x-2)(z-2)]
=(z-2+x-2+y-2)/[(x-2)(y-2)(z-2)]
=(x+y+z-6)/[xyz-2(xy+xz+yz)+4(x+y+z)-8]
=(2-6)/[1-2(-6)+4*2-8]
=-4/13
xy+2z=xy+4-2x-2y=(x-2)(y-2).
同理，yz+2x=(y-2)(z-2),zx+2y=(z-2)(x-2).
4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx),
xy+yz+zx=-6.
(x-2)(y-2)(z-2)=xyz-2(xy+yz+zx)+4(x+y+z)-8=13.
原式=[(x-2)+(y-2)+(z-2)]/(x-2)(y-2)(z-2)
=-4/13.
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