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已知A与B互为相反数 C与D互为倒数,x=%(A+)%3(1%A),y=C的平方D%(D/C+C...

化简x= -(a-2b)-3(1-a)=-a+2b-3+3a=2b+2a-3=2(a+b)-3 a和b互为相反数 所以和为0=2*0-3=-3x=-3y=c2d+d2-[(d/c) +c-1]=(c^2)d+d^2-(d/c)-c+1=c(cd)+d^2-d^2-c+1 c和d互为倒数 积为1 d/c=d乘c的倒数 也就是d 所以为d^2=

∵a与b互为相反数,∴a+b=0,∵c与d互为倒数,∴cd=1,3x-2y=3[3(a-1)-(a-2b)]-2[cd+c(a+b)]=6a+6b-9-2cd-2ca-2cb=6(a+b)-9-2cd-2c(a+b)=6*0-9-2*1-2c*0=-11.

x=3(a-1)-(a-2b) = 3a-3-a+2b = 2a+2b-3 = -3y=c2d+d2-(d/c+c-2)= c+d^2-d^2-c+2= 22x-y/3-3x+2y/6= -x=3

已知a与b互为相反数,c,d互为倒数则,a+b=0 cd=1x=3(a-1)-(a-2b)=3a-3-a+2b=2a+2b-3=2*(a+b)-3=2*0-3=-3y=cd+c(a+b)=1+c*0=1则,3x+2y=3*(-3)+2*1=-9+1=-8

a与b互为相反数所以:a+b=0c与d互为倒数所以:cd=1x=3(a-1)-(a-2b), =2a+2b-3 =2(a+b)-3 =-3cd=1两边同时除以c得出:d/c=1/cy=c^2d+d^2-(d/c+c-1) =cd*c+d^2-(1/c+c-1) =d^2-1/c+1 =(cd^2-1+c)/c =1所以:(2x+y/3)-(3x-2y/6) =3+2/3 =11/3

a+b=0cd=1x=3a-3-a+2b=2a+2b-3=-3y=c*c*d+d*d-(d/c+c-1)=c+d-(d+c-1)=c+d-d-c+1=1代数式(3x-2y)/2-(4x-2y)/3=-11/2+14/3=-33/6+28/6=-5/6

解a,b互为相反数则a+b=0c,d互为倒数则cd=1x=3(a-1)-(a-2b)=3a-3-a+2b=2(a+b)-3=-3y=cd+d-(d/c+c-1)cd=1则c=1/d, =c+d-(d+c-1) =c-c+1 =1∴(2x+y)/3-(3x-2y)/6是这个吧?=(-6+1)/3-(-9-2)/6=-5/3+11/6=1/6

已知a与b互为相反数,a+b=0c与d互为倒数,cd=1;x=3(a-1)-(a-2b)=3a-3-a+2b=2a+2b-3=2(a+b)-3=2(0)-3=-3;y=(c^2)d+d^2-(d/c+c-1)=cd+d-d/c-c+1=c(cd)+[(cd)d-d]/c-c+1=c(1)+[(1)d-d]/c-c+1=c+[d-d]/c-c+1=c+0-c+1=1(3x-2y)/2-(4x-2y)/3=(-9-2)/2-(-12-2)/3=-11/2+14/3=(-33+28)/6=-5/6

题目应是:已知a与b互为相反数,c与d互为倒数,x= -(a-2b)-3(1-a),y=c的平方乘d-c+1,a+b=0 cd=1x= -(a-2b)-3(1-a)= -a+2b-3+3a=2a+2b-3=2(a+b)-3=-3y=c的平方乘d-c+1=c^2d-c+1=c-c+1=13分之2x+y减6分之3x减2y

x= -3 y= 1 答案: 负六分之五

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