初一数学，分式的运算，给出详细过程
\u521d\u4e00\u6570\u5b66\u5206\u5f0f\u8ba1\u7b973/\uff08(x-2)(x+1)\uff09-\uff081/\uff08x-2\uff09\uff09
=3/\uff08(x-2)(x+1)\uff09-\uff08x+1)/\uff08x-2\uff09(x+1\uff09
=(3-x-1)/(\uff08x-2\uff09(x+1\uff09
=(2-x)/\uff08x-2\uff09(x+1\uff09
=-1/(x+1)
2/\uff08a²-1\uff09+1/\uff08a+1\uff09
=2/\uff08a²-1\uff09+(a-1)/\uff08a+1\uff09(a-1)
=(2+a-1)/(\uff08a²-1\uff09
=(a+1)/(\uff08a²-1\uff09
=1/(a-1)
x/(2-x)-(4-x²)/(x²-4x+4)
=-x(x-2)/ (x²-4x+4)-(4-x²)/(x²-4x+4)
=(-x²+2x-4+x²)/(x²-4x+4)
=-2(x-2)/(x²-4x+4)
=-2/(x-2)
=2/(2-x)
3x²+xy-2xy=(x+y)(3x-2y)=0
x+y=0\u62163x-2y=0
\u539f\u5f0f=[(x+y)/(x-y)-4xy/(x+y)(x-y)]\u00f7[(x-y)(x+3y)/(x+3y)(x-3y)]
={[(x+y)²-4xy]/(x+y)(x-y)}\u00f7[(x-y)/(x-3y)]
=[(x²-2xy+y²)/(x+y)(x-y)]\u00d7[(x-3y)/(x-y)]
=[(x-y)²/(x+y)(x-y)]\u00d7[(x-3y)/(x-y)]
=[(x-y)/(x+y)]\u00d7[(x-3y)/(x-y)]
=(x-3y)/(x+y)
x+y\u5728\u5206\u6bcd\uff0c\u4e0d\u7b49\u4e8e0
\u6240\u4ee53x-2y=0
y=3x/2
\u6240\u4ee5\u539f\u5f0f=(x-9x/2)/(x+3x/2)
=(-7x/2)/(5x/2)
=-7/5
绛旓細1锛歺y/(x+y)=1鍙帹鍑(x+y)/xy=1 鍗 1/x+1/y=1,鍚岀悊锛1/y+1/z=2,1/x+1/z=5 涓変釜鏂圭▼瑙ｅ緱锛1/x=2,1/z=3,1/y=-1 1/xyz=-6,xyz=-1/6;2:x/3y=y/(2x-5y)鍙帹鍑2x^2-5xy=3y^2, 锛堜竴锛墄/3y=(6x-15y)/x鍙帹鍑簒^2=18xy-45y^2, 锛堜簩锛夛紙涓锛壝15锛...
绛旓細鍥炵瓟锛氱瓟妗堣鍥锯︹
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绛旓細璇ュ伐绋嬮檺鏈焫澶╁畬鎴 鐢插伐绋嬮槦鍗曠嫭鍋歺-1澶╁畬鎴 涔欏伐绋嬮槦鍗曠嫭鍋歺+2澶╁畬鎴 1/(x-1)+x/(x+2)=1 涓よ竟鍚屼箻浠(x-1)(x+2)寰楋細x+2+x(x-1)=(x-1)(x+2)x+2+x²-x=x²+x-2 x=4 璇ュ伐绋嬮檺鏈4澶╁畬鎴
绛旓細灏嗘柟绋嬪睍寮寰楋細a2+2ab-mab-2mb2 = a2-nab-6b2 鍗 a2+锛2-m锛塧b-2mb2= a2-nab-6b2 寰2-m=-n, -2m=-6 瑙ｅ緱锛歮=3,n=1
绛旓細鍥炵瓟锛氣憼8/(x-3)=2 8=2x-6 x=7 鈶(3-x)/(x+4)=1 3-x=x+4 x=-0.5
绛旓細(1):1/2-3X=3/2X+1 瑙ｏ細鏂圭▼涓よ竟鍚屼箻2寰 锛2涓哄叕鍒嗘瘝锛1-6X=3X+2 ( -6X-3X=2-1 ) 杩欐绉颁负绉婚」 绉婚」鏃惰鏀瑰彉绗﹀彿 瑙ｅ緱X=-1/9 娉細鎷彿鍐呯殑鍙渷鐣 锛2锛夛紙X-1)/4+3=(2X+1)/2 瑙ｏ細鏂圭▼涓よ竟鍚屼箻14寰 锛14涓哄叕鍒嗘瘝锛2锛圶-1)=7(2X+1)( 2X-2=14X+7 ...
绛旓細(x+2)/(x-2)-1=3/(x²-4)(x+2-x+2)/(x-2)=3/(x-2)(x+2)(x鈮2)4=3/(x+2)4x+8=3 4x=-5 x=-5/4 浣犵殑棰樼洰涓嶅娓呮櫚
绛旓細绛旓細姣忎竴椤归兘鍙互杩涜瑁傞」鍒嗚В =1/(2x+1)-1/(2x+2)+1/(2x+2)-1/(2x+3)+1/(2x+3)-1/(2x+4)+1/(2x+4)-1/(2x+5)=1/(2x+1)-1/(2x+5)=4/[(2x+1)(2x+5)]涓棿鍚勯」姝ｈ礋鐩告姷娑堬紝鍓╀綑鏈鍚庝竴椤瑰拰绗竴椤
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