3m-2n乘以2m n-4n乘以3m-2n
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![3m-2n乘以2m n-4n乘以3m-2n](/uploads/image/f/296674-34-4.jpg?t=3m-2n%E4%B9%98%E4%BB%A52m+n-4n%E4%B9%98%E4%BB%A53m-2n)
[|16-m^2|+4×(m-2n)^2]/(m+4)^1/2=0列下来是这样.没错吧?以后发题目尽量不要用语言描述啊,歧义很多读起来也不方便.m不能等于-4.乘过去,|(16-m)^2|+4×(m-
2x^1-(3m+n)x+mn=0,m>n>0x1=[3m+n+√(n^2-2mn+9m^2)]/4={3m+n+√[(n-m)^2+8m^2]}/4>(3m+n+2√2m)/4>(3n+n+2√2n
m和n互为相反数,m+n=0,m=-n,n不等于0,则m也不等于0,mn²+m²n-2n/m=mn(n+m)-2n/(-n)=mn(0)+2=0+2=2
已知m(m-1)-(m²-n)=3,去括号展开,得:m²-m-m²+n=3,所以n-m=3.现在求(m²+n²)/2-mn,通分,变一下形:(m
9M^2-4n^2原式=(3M)^2-(2n)^2=(3M2n)(3M-2n)
因为(M^2-N^2)=(M+N)(M—N)所以原式=[(M+N)/(M-N)]*(M-N)^2/[(MN)^2(M+N)]=(M-N)/(NM)^2
2m/3n*(3n/p)^2/(mn/p)=2m/3n*9n^2/p^2*p/mn=6/p
4乘以(-4分之3)的2n次方乘以8的m次方乘以(1又3分之1)的2n次方乘以(0.125)的m次方+8的n次方除以4的n次方=4×(-3/4)²ⁿ×8^m×(4/3)²
2分之1m乘以mn乘以m减去m乘以n乘以m的二次方加上2分之1m的二次方乘以n乘以m=1/2m×mn×m-mn×m²+1/2m²n×m=1/2m³n-m³n+1
就是证明b平方-4ac>0化简以下可以得到2m^2+(m-n)^2>0这是恒等式所以m>n>0都不需要就可以得到2乘以x的平方+(3m+n)x+mn=0有两个不相等的实数根
3(m^2n+mn)-4(mn-2m^2n)+mn.=3m^2n+3mn-4mn+8m^2n+mn.=11m^2n.
(-2*m*n+2*m+3*n)-(3*m*n+2*n-2*m)-(m+4*n+m*n)=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=-2mn-3mn-mn+2m+2m-m+3n-2n-
(2分之1m+n)乘以(m-2n)=(m+2n)(m-2n)/2=(m²-4n²)/2=m²/2-2n²再问:(m+2n)(m-2n)/2为什么n扩大了两倍第一
(a^2m)^3+(b^n)^3-a^2m×b^n×a^4m×b^2n=a^6m+b^3n-a^6m×b^3n=(a^3m)^2+b^3n-(a^3m^2×b^3n)=3^2+2-(3^2×2)=9+
m=2*3*an=2*7*a2a=10a=5打字不易,
n/(m²+2mn)=n/[m(m+2n)]m/(2n²-mn)=-m/[n(m+2n)]2mn/(m²-4n²)=2mn/[(m+2n)(m-2n)]所以n/