|m-n-2|=-(mn-1); 所以m-n-2=0;m=n+2; mn-1=0; 原式=(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn) =-2mn+2m+3n-3mn-2n+2m-m-4n-mn =-6mn+3m-3n =-6*1+3*2 =-6+6 =0;
已知|m-n-2|与(mn-1)的平方 互为相反数,求(-2mn+2m+3n)-(3mn+2n-3m)-(m+4n+mn
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