若2x-y=2m和x 3y=m-1满足x
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(2m^2+m-2)x+(m^2-m)y+4m-1=0和直线2x-3y=5的斜率分别是(2m^2+m-2)/(m-m^2)和2/3因为直线(2m^2+m-2)x+(m^2-m)y+4m-1=0和直线2
x3y+2x2y2+xy3=xy(x2+2xy+y2)=xy(x+y)2,∵x+y=5,∴(x+y)2=25,x2+y2+2xy=25,∵x2+y2=13,∴xy=6,∴xy(x+y)2=6×25=1
m^x+y=m^x×m^y=2×3=6m^3x+2y=m^3x×m^2y=(m^x)^3×(m^y)^2=8×9=72
已知x^(3m)=2y^(2m)=3(x^(2m))^3+(y^m)^6-(x^2*y)^3m*y^m=x^6m+y^6m-x^6my^4m=(x^3m)^2+(y^2m)^3-(x^3m)^2*(y
依题意得不等式组m^2-m≠0m=2解得m=2
用第一个式子减去第二个式子3x+5y-x-2y=m-4-m2x+3y=-4由题目得知x+y=1连立以上两个方程,解得x=7,y=-6将x、y代回式子中,得到m=-5
2x+y=2m-1①x+2y=m②①+②得3x+3y=3m-1x+y=m-1/3①-②得x-y=m-1∵x+y>0,x-y0,m-11/3,m
∵x+y=4,∴(x+y)2=16,∴x2+y2+2xy=16,而x2+y2=14,∴xy=1,∴x3y-2x2y2+xy3=xy(x2-2xy+y2)=14-2=12.
(2x4-4x3y-x2y2)-2(x4-2x3y-y3)+x2y2=2x4-4x3y-x2y2-2x4+4x3y+2y3+x2y2=2y3,因为化简的结果中不含x,所以原式的值与x值无关.
原式=(x^4-2x²y²+y^4)+6xy(x²+2xy+y²)-2xy(x+y)=(x²-y²)²+6xy(x+y)²
第一题:两式相加得x+y=4m+3+m-4即x+y=5m-1∴5m-1=3m=五分之四第二题:设该商品进价x元,定价y元由题意可得y-x=25按定价的七五折销售该商品8件与将定价降低15元销售该商品1
联列两直线方程y=1/2x,y=-x+m得点M(2/3m,1/3m)对二次曲线求导y'=2x+p,令y’=0,将顶点M代入得p=-4/3m,再代入二次曲线函数得q=1/3m+4/9m^2将曲线方程y=
已知x+y=5,xy=3,代数式x3y-2x平方y平方+xy3=xy(x²-2xy+y²)=xy(x-y)²=3×[(x+y)²-4xy]=3×(25-12)=
3x+2y=m2x-y-1=2m2x-y-1=2(3x+2y)2x-y-1=6x+4y4x+5y=-1x-y=29y=-9y=-1x=13x+2y=1m=1
应该是X3y-2x2y2+xy3原式=x3y-2x2y2+xy3=xy(x2-2xy+y2)=xy(x-y)2=17/36*6=17麻烦采纳,谢谢!
你好,解决方法如下:记方程组为①②①*2得,2mx+6y=4③②*m得,2mx+m(m-1)y=m²④④-③得,[m(m-1)-6]y=m²-4⑤一次方程有无穷解的情况就是0*x=
x+y=4,xy=2后者平方后二式相加再加后者平方
x3y+xy3=xy(x^2+y^2)=(√3-√2)(√3+√2)((√3-√2)^2)+(√3-√2)^2)=1*(3-2√6+2+3+2√6+2)=10
和是单项式则两个是同类项所以x和y的次数分别相等所以m+5=22=n所以m=-3,n=2所以原式=2^(-3)=1/8
∵x-y=l,xy=2,∴x3y-2x2y2+xy3=xy(x2-2xy+y2)=xy(x-y)2=2×1=2.