若a-b=2,a-c=2分之1,求(b-c)³-(b-c)+4分之9的值
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a+b/ab+b+c/bc+c+a/ac=1/b+1/a+1/c+1/b+1/a+1/c=2(1/a+1/b+1/c)∵1/a+1/b+1/c=-2∴所求=2×(-2)=-4再问:看补充的问题!再答:
∵a(b分之1+c分之1)+b(c分之1+a分之1)+c(a分之1+b分之1)+3=a(1/b+1/c)+a/a+b(1/c+1/a)+b/b+c(1/a+1/b)+c/c=a(1/a+1/b+1/c
原式a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+2=a/b+a/c+b/a+b/c+c/a+c/b+2=(b+c)/a+(a+c)/b+(a+b)/c+2∵a+b+c=0且abc
已知a:b=2分之1=1:2=5:10b:c=5:6=10:12所以a:b:c=5:10:12
解令a/5=b/7=c/8=k∴a=5kb=7kc=8k∵3a-2b+c=-9∴15k-14k+8k=-9∴9k=-9∴k=-1∴2a+4b-3c=10k+28k-24k=14k=-14
a是1」4.b是-1」4.c是1」2再问:过程
分之a等于3分之1即:b=3a代入:3a+2b=2分之3得:b+2b=3/23b=3/2b=1/2a=b/3=(1/2)/3=1/6所以:a=1/6b=1/2
(1)cos(A+C)=2[cos((A+C)/2)]^2-1=-1/3cosB=-cos(A+C)=1/3(2)向量BA乘向量BC=c×a×cosB
正弦定理因为:cosB/cosC=-b/2a+c=-sinB/(2sinA+sinC)所以:2cosBsinA+cosBsinC=-sinBcosC整理得:(2cosB+1)sinA=0cosB=-1
若B分之A等于5分之7,c分之B等于2分之3,则B分之A等于15分之21,c分之B等于14分之21,所以A=21k,B=15k,C=10kB+C分之A-B=(21k-15k)/(15k+10k)=6/
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=a(b+c)/(bc)+b(a+c)/(ac)+c(a+b)/(ab)=a^2(b+c)/(abc)+b^2(a+c)/(abc)+c
1.令a/3=b/4=c/5=t则:a=3t,b=4t,c=5t(2a+b-c)/(a-b+2c)=5t/9t=5/92.令(a+b-c)/c=(a-b+c)/b=(b+c-a)/a=n∴a+b-c=
A=十八分之3B=十八分之十二C=十八分之五
设a+b=x,b+c=y,a+c=z,那么x+y+z=2(a+b+c),2(a+b+c)/(a+b)+2(a+b+c)/(b+c)+2(a+b+c)/(a+c)=(x+y+z)/x+(x+y+z)/y
原式=b/(a-b+c)-(2a+c)/(a-b+c)+(b-c)/(a-b+c)=(b-2a-c+b-c)/(a-b+c)=-2(a-b+c)/(a-b+c)=-2
题目没讲清楚啊,如果是(a+b-c)/(a-b)/c=(1+2-3)/(1-2)/3=0/(1-2)/3=0,若是(a-b)/c/(a+b-c)=(1-2)/3/(1+2-3)=(1-2)/3/0,则
(1)(a-b)(a-c)分之2a-b-c+(b-c)(b-a)分之2b-c-a+(c-b)(c-a)分之2c-a-b=1/(a-b)+1/(a-c)+1/(b-c)+1/(b-a)+1/(c-b)+
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B-A=1/2A=B-1/2B-C=7/18C=B-7/18代入A+B+C=10/9B-1/2+B+B-7/18=10/93B=2B=2/3A=B-1/2=1/6C=B-7/18=5/18
证:a/b=(a-c)/(c-b)b(a-c)=a(c-b)ab-bc=ac-abac+bc=2abc(a+b)=2ab(a+b)/(ab)=2/c1/a+1/b=2/c等式成立.