已知x y都是正数若x 2y=3求1 x 1 y的最小值
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/21 00:45:12
x2y+xy2-x-y=xy(x+y)-(x+y)=(x+y)(xy-1)∵x+y=-5,xy=7,∴原式=-5×(7-1)=-30.
(x+y)(xy)=x^2y+xy^2=-8原式=-7
(1)4ab+8-2b2-9ab-6=-2b2-5ab+2(2)原式=3x2y-2x2y+6xy-3x2y+xy=-2x2y+7xy,当x=-1,y=-2时,原式=-2×(-1)2(-2)+7×(-1
都是1.5 你的题目看错了吗?y上的为1如果我没错的话你们根本做不了请采纳我哦.
x+y+xy=9x+y=9-xyx^2y+xy^2=20xy(x+y)=20xy(9-xy)=20xy^2-9xy+20=0(xy-4)(xy-5)=0xy=4或xy=5x+y=5或x+y=4x^2+
由已知:xy+x+y=17,xy(x+y)=66,可知xy和x+y是方程t2-17t+66=0的两个实数根,得:t1=6,t2=11.即xy=6,x+y=11,或xy=11,x+y=6.x2+y2=(
因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.
①x2y+xy2=xy(x+y)=1×3=3;②x2+y2=(x+y)2-2xy=32-2×1=7.
原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.
x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10
那个2是平方吧?可以用^代替原式=x^y+xy^=xy(x+y)=-3*6=-18
是不是求:5x²y-[2x²-(3xy-xy²)-3x²]-2xy²-y²再问:是再答:已知是不是(x+3)²+|x+y+10|=
(x+y)(x-y)-y^2+(x-y)^2-(6x^2y-2xy^2)/(2y)=X^2-y^2-y^2+X^2+y^2-2xy-3x^2+xy=-x^2-y^2-xy=-(x^2+y^2+xy-3
1)x,y都是正数12=3x+2y≥2√(6xy)√xy≤√6xy≤62)x+2y=31/x+1/y=1/x+2/3-x=(3+x)/x(3-x)=(x+3)/[6(3+x)-x(3+x)-18]=1
∵x>0,y>0,∴12=3x+2y≥23x×2y,化为xy≤6,当且仅当3x=2y,3x+2y=12,即x=2,y=3时取等号.∴xy的最大值为6.
最大值14程序如下:p=0q=0for(i=1,i
因为x、y都是正数,则:x+4y≥4√(xy)设:√(xy)=t,则:xy=4y+x+5≥4√(xy)+5即:t²≥4t+5t²-4t-5≥0t≤-1或t≥5因为:t=√(xy)≥
由题意得:3C=A+B=8x2y-6xy2-3xy+7xy2-2xy+5x2y=13x2y+xy2-5xy,∴C=13x2y+xy2−5xy3,故:C-A=13x2y+xy2−5xy3-(8x2y-6
∵x2-y2=xy,∴原式=x2y2+y2x2=x4+y4x2y2=(x2−y2)2+2x2y2x2y2=3x2y2x2y2=3.再问:先化简2a+1/a²-1÷a²-a/a