(x-y)的2m次方*(y-x)的2x-1次方
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稍等一下再问:嗯再答:解x^3m=4y^3n=5(x^m)³=4∴x^m=³√4(y^n)³=5∴y^n=³√5(x^2m)³+(y^n)³
=七分之一x的m-1次方y的m+1次方(21xy+1)说明:把第二项作为公因式提取即可再问:打错了,这才是题目:多项式3x的m次方y的n+2次方+七分之一x的m-1次方y的n+1次方分解因式的结果是多
x^3y^(m-1)x^(n+m)y^(2n+2)=x^(3+n+m)y^(m-1+2n+2)=x^(m+n+3)y^(m+2n+1)=x^9y^9m+n+3=9m+2n+1=9解得m=4n=24m-
(x-y)的m次方×(y-x)的2m次方+(y-x)的3m次方=(x-y)的(m+2m)-(x-y)的3m次方=0
x^(m-n)*x^(2n+1)=x^11.(m-n)+(2n+1)=11.m+n=10y^(m-1)*y^(4-n)=y^5.(m-1)+(4-n)=5.m-n=2(m+n)(m-n)=20m=6n
5m(x-y)²+10n(y-x)³=5m(x-y)²-10n(x-y)³=5(x-y)²[m-2n(x-y)]=5(x-y)²(m-2nx
x³+3x-4=(x-1)(x^2+x+4)再问:可以给个过程吗?谢谢再答:定理:当一个只含有一个未知数的多项式当未知数为a时多项式值为0则多项式必有因式(x-a)当x=1时,原式=0∴x&
【6x的n次方y的m次方(x的平方-3xy)-(-3x的n次方y的m次方)的2次方]除以3x的n次方y的m-n次方=[6x的(n+2)次方y的m次方-18x的(n+1)次方y的(m+1)次方-9x的2
根据叙述列出的式子为:x^(m-2)*y^2+m*x^(m-2)*y+n*x^3*y^(m-3)-2*x^(m-3)*y+m+n因为此式化简后是个四次三项式,而现在式子为六项,那么肯定要合并同类项,所
(x-y)(z+y-x+y+2y)÷4y=(x-y)(z-x+4y)÷4y{(x+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y=(x-y)(x+y-x+y+2y)÷4y=(x-y)(4
(x+y)²(2x+2y)²(3x+3y)²=(x+y)²【2(x+y)】²【3(x+y)】²=36(x+y)的6次方=36m的6次方
x=3^m-1y=2-9^m3^m=x+19^m=(3^m)^2=(x+1)^2=x^2+2x+1y=2-(x^2+2x+1)y=-x^2-2x+1
x=2^m,y=4^(3m)y=4^(3m)=(2^2)^(3m)=(2^(3m))^2=(2^m*2^m*2^m)^2=(x^3)^2=x^6
5m(x-y)^2+10n^2(y-x)^3=5(x-y)^2(m+2n^y-2n^x)那个^2是指……的2次方
2^m=x-1则4^m=(x-1)^2=y-3y=x^2-2x+4后一个题目不完整若x^m*x^2m=x^m=a则x^9m=a^3
(m+n)x^ny^(m-2)(3xy^2+5x^2y)=21x^my^(n+1)+35x^(m+1)y^n=7x^(m-1)y^(n-1)(3xy^2+5x^2y)m+n=7n=m-1m-2=n-1
计算:x·x²·x的n次方-x的n+3次方=x的n+3次方-x的n+3次方=0y·y的m+1次方·y的m-1次方-y的2m+1次方.=y的(2m+1)次方-y的(2m+1)次方=0
x^(n-m)*x^(2n+1)=x^11==>(n-m)+(2n+1)=11(1)y^(m-1)*y^(4-n)=y^5==>(m-1)+(4-n)=5(2)由(1)、(2)式解方程组得m=8,n=
解由已知得原式=x^6m+y^3n-x^6m*y^3n=16+5-16*5=-59同底数幂相乘,底数不变,指数相加即:a^m×a^n=a^(m+n)(a^m)^n=a^mn(这个忘了咋说的了.)
即2x(2+m)y^2+6x^3y^4=2x^5y^2-6x^3y^n所以2+m=5,4=n则m=3,n=4