作业帮快来教教我已知a与b互为相反数,c与d互为倒数,x= -(a-2b)-3(1-a),y=c2d+d2-[(d/c) +c-1]求(2x+y/3)–(3x-2y/6)的值注：y=c2d+d2-[(d/c) +c-1]中的c2d为c的平方乘d,d2为d的平方
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![快来教教我已知a与b互为相反数,c与d互为倒数,x= -(a-2b)-3(1-a),y=c2d+d2-[(d/c) +c-1]求(2x+y/3)–(3x-2y/6)的值注：y=c2d+d2-[(d/c) +c-1]中的c2d为c的平方乘d,d2为d的平方](/uploads/image/z/7653781-37-1.jpg?t=%E5%BF%AB%E6%9D%A5%E6%95%99%E6%95%99%E6%88%91%E5%B7%B2%E7%9F%A5a%E4%B8%8Eb%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%2Cc%E4%B8%8Ed%E4%BA%92%E4%B8%BA%E5%80%92%E6%95%B0%2Cx%3D+-%28a-2b%29-3%281-a%29%2Cy%3Dc2d%2Bd2-%5B%28d%2Fc%29+%2Bc-1%5D%E6%B1%82%282x%2By%2F3%29%E2%80%93%283x-2y%2F6%29%E7%9A%84%E5%80%BC%E6%B3%A8%EF%BC%9Ay%3Dc2d%2Bd2-%5B%28d%2Fc%29+%2Bc-1%5D%E4%B8%AD%E7%9A%84c2d%E4%B8%BAc%E7%9A%84%E5%B9%B3%E6%96%B9%E4%B9%98d%2Cd2%E4%B8%BAd%E7%9A%84%E5%B9%B3%E6%96%B9)
快来教教我已知a与b互为相反数,c与d互为倒数,x= -(a-2b)-3(1-a),y=c2d+d2-[(d/c) +c-1]求(2x+y/3)–(3x-2y/6)的值注：y=c2d+d2-[(d/c) +c-1]中的c2d为c的平方乘d,d2为d的平方
快来教教我
已知a与b互为相反数,c与d互为倒数,
x= -(a-2b)-3(1-a),y=c2d+d2-[(d/c) +c-1]
求(2x+y/3)–(3x-2y/6)的值
注：y=c2d+d2-[(d/c) +c-1]中的c2d为c的平方乘d,d2为d的平方
快来教教我已知a与b互为相反数,c与d互为倒数,x= -(a-2b)-3(1-a),y=c2d+d2-[(d/c) +c-1]求(2x+y/3)–(3x-2y/6)的值注：y=c2d+d2-[(d/c) +c-1]中的c2d为c的平方乘d,d2为d的平方
jason_wujch - 助理 二级 11-10 20:58的是对的 我给你详解
已知a与b互为相反数,c与d互为倒数,
x= -(a-2b)-3(1-a),y=c2d+d2-[(d/c) +c-1]
求(2x+y/3)–(3x-2y/6)的值
注：y=c2d+d2-[(d/c) +c-1]中的c2d为c的平方乘d,d2为d的平方
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
=-3
x=-3
y=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
=c-c+1
=1
x=-3 y=1代入式子 得
(-6+1/3)-(-9-1/3)
=3+2/3
=11/3
已知a与b互为相反数，c与d互为倒数，
x= -(a-2b)-3(1-a),y=c2d+d2-[(d/c) +c-1]
===>x=-3a-3+3a=-3,y=c+d^2-d^2-c+1=1
===> (2x+y/3)–(3x-2y/6)=-x+2y/3=3+2/3=11/3
a与b互为相反数，c与d互为倒数,
a+b=0,cd=1,c/d=-1,d/c=-1,
x= -(a-2b)-3(1-a)
=-a+2b-3+3a
=2a+2b-3
=2(a+b)-3
=2*0-3
=-3,
y=c^2d+d^2-[(d/c) +c-1]
=c^2d+d^2-d/c-c+1
=c*cd+d^2-1-c...
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a与b互为相反数，c与d互为倒数,
a+b=0,cd=1,c/d=-1,d/c=-1,
x= -(a-2b)-3(1-a)
=-a+2b-3+3a
=2a+2b-3
=2(a+b)-3
=2*0-3
=-3,
y=c^2d+d^2-[(d/c) +c-1]
=c^2d+d^2-d/c-c+1
=c*cd+d^2-1-c+1
=c*1+d^2-c
=d^2.
(可能是题目抄写错误,不能解得y的值.)x,y的值确定后,代数式(2x+y/3)–(3x-2y/6)的值也就能确定下来了.
收起
化简x,y,a与b互为相反数，c与d互为倒数,化简x,y
x=-(a+2a)-3(1-a)=-3,
y=c^2*(1/c)+(1/c)^2-[(1/c)^2+c-1]=1
(2x+y/3)–(3x-2y/6)=-6+1/3+9+1/3=11/3
(2x+y)/3–(3x-2y)/6=-5/3+11/6=1/6
化简 x=3b-3-3b=-3 y=c+1/c2-(1/c2+c-1)=1
即x=-3 y=1代入下面计算得
(2x+y/3)–(3x-2y/6)=(-6+1/3)-(-9-1/3)
=3+2/3