1.计算1(m+2n)(m-2n);2(2n+5)(n-3);3(x+2y)²;4(2x+b)(3x+d);
1.计算1(m+2n)(m-2n);2(2n+5)(n-3);3(x+2y)²;4(2x+b)(3x+d);
已知,单项式M,N满足2x(M+3x)=6x²y²+N,求M,N
计算(1)(3m-4n)(3m+4n)-(2m-2n)(2m+n) (2)(x+5y)²-(x-5y)
计算:1.[(m+n)(n-m)-(m+n)^2+3m(m+n)]除以(-2M),其中m=-4,n=2 2.(x+y)(
几道分式计算.【1】[1/(m+n)^2-1/(m-n)^2]÷(1/m+n-1/m-n)【2】2/3x-(2/x+y)
已知|m-2|+(3-3n)²=0,化简2x^m-n+1 y^3-6y^m+n x²
提取公因式法 有分6(m-n)³-12(n-m)²2(y-x)²+3(x-y)mn(m-n
函数y=(m-3)x+n-2的图象如图所示,化简代数式:|m-n|-(根号n²-4n+4)-|m-1|
已知(m-2)x^|m-1|-(n+3)y^n²-8=1是关于x,y的一元一次方程,且m,n满足{ma+nb=5
集合M={X=3m+1,m∈N*},N={y/y=5n+2,n∈N*},则M∩N=?
分式计算题2道 3x/(x-3)² - x/3-x m+2n/n-m + m/n-m - 2m/n-m
计算:(1)(m-n)^2*(n-m)^5*(m-n)^5 (2)(-x)^2*x^3+2x^3*(-x)^2-x*x^