Cho \(x^3+y^3=2\). Tim Max x +y
cho x+y+4=0
Tim max B=2(x^3+y^3)+3(x^2+y^2)+10xy
Ta có B=\(2\left(x+y\right)\left(x^2-xy+y^2\right)+3x^2+3y^2+10xy\)
\(B=-8x^2+8xy-8y^2+3x^2+3y^2+10xy\)
\(-B=5x^2-18xy+5y^2>=\frac{5}{2}\left(x+y\right)^2-18\left(\frac{x+y}{2}\right)^2=40-72\)=-32
hay b>=32 dấu bằng xảy ra tự tính
cho x+y=5 tim max A=x^4+y^4-4(x^3+y^3)-20(x^2+y^2)-2x^2y^2+xy
1)Tim MAX cua A= (6x^2-2x+1)/ x^2
2)tim MIN va MAX C= (3-4x)/(X^2+1)
3) Tim MIN va MAX P = x^2+y^2
biet giua x va y co moi quan he nhu sau : 5x^2+8xy+5y^2=36
4)tim MAX Q = -x^2-y^2+xy+2x+2y
Cho 2 so thuc x , y thoa man x^2 + 4y^2 = 8
Tim max cua M=y ( 2 x - 3 y)
cho x^2+y^2+z^2=3 a cmr x^2y+y^2z+z^2x=<2+xyz b tim max min x/y+2+y/z+2+z/x+2
Cho x>=0,y,z<=2,x+y+z=3.tim min,max x^4+y^4+z^4+12(1-x)(1-y)(1-z)
Đặt \(\hept{\begin{cases}a=x-1\\b=y-1\\c=z-1\end{cases}}\)\(-1\le a,b,c\le1\) và \(a+b+c=0\)
\(T=(a+1)^4+(b+1)^4+(c+1)^4-12abc\)
\(=a^4+b^4+c^4+4(a^3+b^3+c^3)+6(a^2+b^2+c^2)+4(a+b+c)+3-12abc\)
Từ \(a+b+c=0\Rightarrow a^3+b^3+c^3=0\). Do đó:
\(T=a^4+b^4+c^4+6(a^2+b^2+c^2)+3\ge3\)
Xảy ra khi \(a=1;b=-1;c=0\)
cho x,y thoa man: x^2+y^2= x+y .tim MIN ,MAX cua B=x-y
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tìm Min ( x^2 + y^2 ) / xy đk x>= 2y; x,y dương? | Yahoo Hỏi & Đáp
Tìm Min:
\(x=x^2+y^2-y\)
\(\Rightarrow B=\left(x^2+y^2-y\right)-y=x^2+\left(y^2-2y+1\right)-1=x^2+\left(y-1\right)^2-1\ge-1\)
Tìm Max:
\(y=x^2+y^2-x\)
\(\Rightarrow B=x-\left(x^2+y^2-x\right)=-y^2-\left(x^2-2x+1\right)+1=-y^2-\left(x-1\right)^2+1\le1\)
cho 0 <= x,y <=1 va x+y=3xy. tim min, max cua P= x^2 + y^2 -4xy
1 tim MAX cua (x+z)(y+t) biet x^2+y^2+z^2+t^2=1
2 tim MAX cua (x+z)(y+t) biet x^2+y^2+2z^2+2t^2=1