一道简单数学题，很急很急
\u4e00\u9053\u5f88\u7b80\u5355\u7684\u6570\u5b66\u9898\u9009B \u627e\u4e00\u79cd\u7279\u6b8a\u60c5\u51b5\uff0c\u5373\u83f1\u5f62
\u8bbe\u5c0f\u5b69\u4eba\u6570\u4e3ax\u4eba\uff0c\u5219\uff1a
\u7532\u603b\u82b1\u8d39\uff1a240+120x
\u4e59\u603b\u82b1\u8d39\uff1a\uff08x+1\uff09*240*60%=144*\uff08x+1\uff09
1\uff09240+120*x=144\uff08x+1\uff09
\u89e3\u5f97\uff1ax=4
\u5c0f\u5b69\u4e3a4\u4eba\u65f6\uff0c\u4e24\u5bb6\u6536\u8d39\u4e00\u6837\u591a
2\uff09240+120*x<144\uff08x+1\uff09
\u89e3\u5f97\uff1ax>4
\u6545\u5f53\u5c0f\u5b69\u6570\u5927\u4e8e4\u4eba\u65f6\uff0c\u7532\u65c5\u884c\u793e\u66f4\u4e3a\u4f18\u60e0
\u5f53\u5c0f\u5b69\u6570\u7b49\u4e8e4\u4eba\u65f6\uff0c\u4e00\u6837
\u5f53\u5c0f\u5b69\u6570\u5c0f\u4e8e4\u4eba\u65f6\uff0c\u4e59\u65c5\u884c\u793e\u66f4\u4e3a\u4f18\u60e0
=(x+1)/x(x+1)(x-1) + (x-1)/x(x+1)(x-1) + 1/(x+1)(x+2)
=[(x+1)+(x-1)]/x(x+1)(x-1) + 1/(x+1)(x+2)
=(2x)/x(x+1)(x-1) + 1/(x+1)(x+2)
=2/(x+1)(x-1) + 1/(x+1)(x+2)
=[2(x+2)]/(x+1)(x-1)(x+2) + (x-1)/(x+1)(x-1)(x+2)
=[2(x+2)+(x-1)]/(x+1)(x-1)(x+2)
=(3x+3)/(x+1)(x-1)(x+2)
=[3(x+1)]/(x+1)(x-1)(x+2)
=3/(x-1)(x+2)
1/(x＾2 - x) + 1/(x＾2 + x) + 1/(x＾2 +3x + 2)
=1/(x-1)x+1/(x+1)x+1/(x+1)(x+2)
=[（x+2)(x+1)+(x-1)(x+2)+x(x-1)]/[x(x+1)(x-1)(x+2)]
=[x²+3x+3+x²+x-2+x²-x]/[x(x+1)(x-1)(x+2)]
=[3x²+3x+1]/[x(x+1)(x-1)(x+2)]
=1/x（x-1） + 1/x（x+1）+ 1/ （x+1）（x+2）
=（x+1+x-1）/x（x+1）（x-1） +1/ （x+1）（x+2）
=2/（x+1）（x-1） +1/ （x+1）（x+2）
=【2（x+2）+（x-1）】/ （x+1）（x-1）（x+2）
=3（x+1）/（x+1）（x-1）（x+2）
=3/（x-1）（x+2
1/(x＾2 - x) + 1/(x＾2 + x )+ 1/(x＾2 +3x + 2)=1/[x*(x-1)]+1/[x*(x+1)]+1/[(x+1)^2+(x+1)]=1/[x*(x-1)]+1/[x*(x+1)]+1/[(x+1)*(x+2)]=[(x+1)*(x+2)+(x-1)*(x+2)+x*(x-1)]/[x*(x-1)*(x+1)*(x+2)]=3x*(x+1)/[x*(x-1)*(x+1)*(x+2)]=3/[(x-1)*(x+2)]=3/(x^2+x-2)
式子是不是1/(x＾2 - x )+ 1/(x＾2 + x )+ 1/(x＾2 +3x + 2)
1/(x＾2 - x )+ 1/(x＾2 + x )+ 1/(x＾2 +3x + 2)
=1/[x(x-1)] + 1/[x(x+ 1)] + 1/[(x +1)(x + 2)]
=[(x +1)(x + 2)]/[x(x-1)(x +1)(x + 2)]+[(x -1)(x + 2)]/[x(x-1)(x +1)(x + 2)]+[x(x-1)]/[x(x-1)(x +1)(x + 2)]
=[(x +1)(x + 2)+(x -1)(x + 2)+x(x-1)]/[x(x-1)(x +1)(x + 2)]
=(3x^2+x)/[x(x-1)(x +1)(x + 2)]
是x的平方分之一还是x分之一的平方？没有括号之类的运算符么？
绛旓細瑙佷笅鍥撅細鏂滃潯涓1:1锛岃〃绀哄潯搴︿负1:1銆傚嵆 I=h:L=1:1銆備篃灏辨槸h=L 鑰屽鏋渉=L锛屽垯璇存槑杩欎釜鏂滃潯缁勬垚鐨勪笁瑙掑舰鏄瓑鑵扮洿瑙掍笁瑙掑舰銆傚涓嬪浘锛欰C=BC 鍧￠暱涓篈B锛屽垯鏍规嵁鍕捐偂瀹氱悊鏈夛細AB²=AC²+BC²AB=鈭(AC²+BC²)=鈭(3²+3²)=鈭(2脳3²)=3鈭2...
绛旓細3棰 璁句箼杞﹁浜哫鍗冪背 1/5X+12=1/4(300-X)1/5X+12=75-1/4X 1/5X+1/4X=75-12 0.45X=63 X=140
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绛旓細S=鈭歔p(p-a)(p-b)(p-c)]p=(a+b+c)/2 杩欓噷锛歱=(2+a+2a)/2=3a/2+1 S=鈭歔p(p-a)(p-b)(p-c)] =鈭(3a/2+1)(a/2+1)(1-a/2)(3a/2-1)=鈭(9a^2/4-1)(1-a^2/4)鐪嬫牴鍙蜂笅鐨勶細璁撅細a^2/4=t y=(9t-1)*(1-t)=-9t^2+10t-1 y鍙栨渶澶!鍗虫姏鐗╃嚎鐨...
