tim min y= (x^4+x^2+5)/(x^4+2x+1)
Tim Min B = \(\frac{x^4+x^2+5}{x^4+2x^2+1}\)
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
choP=(1/(x-2)-x^2/(8-x^3)*(x^2+2x+4)/(x+2)0/1/(x^2-4) tim DKXD va rut gon b tim Min p c tim x nguyen de p chia het cho x^2+1
tim min cua x4+y4x3-y3+2x2y2+2xy.(x2+y2)+13xy
voi x+y=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 \(a=1;b=-1;c=0\)
Tìm min max của
1,y=sin3 x+cos3 x
2,y=x2 -1/x4 -x2 +1
3,y=4√(x2 -2x+5) +x2 -2x+3
Tim min
P=(x^2 + 1)(y^4 + 1) biết x + y = căn 10
cho x+y=2x^5
va x^4+x^2y^2+y^4=xy(x^2+y^2)+1 tim x va y
Tim x,y:
|x-4|+3|3x-2|-2|2-x|=x+1
||2x-3|-3/4|=5/6
|1/2-|3-x||=2x-1
|x-2007|+2004|y-2008|\(\le\)0
3|x-2y|+4|y+1/2|=0