分式混合运算(习题及答案).pdf

该【分式混合运算(习题及答案) 】是由【青山代下】上传分享，beplayapp体育下载一共【13】页，该beplayapp体育下载可以免费在线阅读，需要了解更多关于【分式混合运算(习题及答案) 】的内容，可以使用beplayapp体育下载的站内搜索功能，选择自己适合的beplayapp体育下载，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此beplayapp体育下载到您的设备，方便您编辑和打印。:..分式混合运算****题及答案)混合运算(题)例1:混合运算:解:原式可以化简为:frac{4-x}{x-2} imes rac{x+2-x^2}{12}$$frac{-(x-4)}{(x-2)(x+4)}$$例2:先化简,然后在的范围内选取一个合适的整数$x$:先化简原式::..frac{x(x+1)}{(x-1)(x-1)} imes rac{x+1}{2x}$$frac{1}{2}$$由于,且$x$为整数,因此使原式有意义的$x$的值为$-2$,$-1$或$2$。代入计算可得:当$x=2$时,原式为$-2$。:化简原式:frac{x-y}{x+2y} imes rac{2x+4y}{1}$$:..frac{2(x-y)}{x+2y}$$化简原式:frac{ rac{a}{a-1}-1}{(a-1)^2} imes(a+1)$$frac{a-2}{(a-1)^2}$$化简原式:frac{2a-2ab}{a^2-b^2} imes rac{a+b}{a+b}$$frac{2a-2ab}{(a-b)(a+b)} imes rac{a+b}{1}$$frac{2(1-b)}{a-b}$$:.. rac{1}{y(y+1)}$$化简原式:frac{y-1- rac{8}{y-1}}{y(y+1)} imes rac{y(y+1)}{1}$$frac{(y-1)^2-8}{y(y+1)^2}$$化简原式:frac{(a-b)^2}{b} imes rac{a-b}{1}-1$$frac{(a-b)^3}{b}-1$$化简原式::..frac{(x-2)^2}{x(x-1)} imes rac{x-1}{x+2}$$frac{(x-2)^2}{x(x+2)}$$7)$$ rac{2}{(x-1)^2}- rac{1}{(x-1)^2(x+1)}$$化简原式:frac{2(x+1)-1}{(x-1)^2(x+1)}$$frac{2x+1}{(x-1)^2(x+1)}$$化简原式:frac{3-x}{2(x-2)} imes rac{x-2}{5}- rac{5}{x-3}$$frac{(x-3)(x-1)}{2(x-2)5}- rac{5}{x-3}$$:..frac{x^2-4x+7}{10(x-2)(x-3)}$$3x}$$化简原式:frac{(x-1)(x-2)}{(x+1)(x-3)}- rac{5}{x(x-3)}$$frac{x^2-3x-2}{x(x-3)(x+1)(x-3)}- rac{5(x+1)}{x(x-3)(x+1)(x-3)}$$frac{x^2-3x-2-5x-5}{x(x-3)(x+1)(x-3)}$$frac{x^2-8x-7}{x(x-3)(x+1)^2}$$10)$$ rac{1}{(x-1)(x+1)}- rac{1}{x(x-1)}$$化简原式::..frac{x-(x-1)}{x(x-1)(x+1)}$$frac{1}{x(x+1)}$$11)$$ rac{2}{x+y}- rac{1}{y-x} imes rac{y^2-x^2}{11}$$化简原式:frac{2(y-x)}{(y-x)(x+y)}- rac{y+x}{11(x+y)}$$frac{y-x-2}{11(x+y)}$$:1)先化简,再求值: rac{1}{x+2}$,其中$x=3-1$。化简原式::..frac{(x+1)^2}{3x+2} imes rac{x+2}{1}$$frac{(4)^2}{11} imes rac{5}{1}$$frac{80}{11}$$2)先化简,再求值:$ rac{ rac{5x+3y}{2}}{22x-yy-x^2y-xy}+ rac{2}{2x+2y}$,其中$x=3+2$,$y=3-2$。化简原式:frac{5x+3y}{4(11x-y-xy)}+ rac{1}{x+y}$$frac{5(3+2)+3(3-2)}{4(11(3+2)-(3-2)-(3+2)(3-2))}+ rac{1}{3+2+3-2}$$frac{25}{38}+ rac{1}{7}$$frac{209}{266}$$:..3)先化简,然后在的范围内选取一个合适的整数$x$代入求值: rac{2x+1}{x-1}$。化简原式:frac{x^2+x+1}{(x-1)^2(x-2)} imes rac{x-1}{2x+1}$$代入$x=-2$:frac{3}{20}$$4)已知$A= rac{x^2+2x+1}{x^2-1}- rac{x}{x-1}$,求$A$当,且$x$为整数时的值。化简$A$:A= rac{(x+1)^2}{(x+1)(x-1)}- rac{x}{x-1}$$:..时,即,且$x$为整数时,$A$的值为:frac{x+2}{x-1}= rac{x-1+3}{x-1}=1+ rac{3}{x-1}$$5)先化简,然后在的范围内选取一个合适的整数$x$代入求值: rac{x^2+x-2}{x-1}$。化简原式:frac{(x-1)^3(x^2+x-2)}{(x-1)^3(x^3-3x^2+3x-1)}$$frac{x^2+x-2}{x^3-3x^2+3x-1}$$代入$x=2$:frac{2^2+2-2}{2^3-3(2)^2+3(2)-1}= rac{2}{3}$$:..6)$-1

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