tim min A= (x+2)^4 + (x-2)^4
cho M=((x^2-1)/(x^4-x^2+1)-1/(x^2+1))(x^4+(1-x^4)/(1+x^2)) a) rut gon b)tim min
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 ( x-3) ^2 + ( x+4 ) ^2
Ta có (x-3)2 và (x+4)2 luôn lớn hơn hoặc bằng không
muốn (x-3)2+(x+4)2 nhỏ nhất thì (x-3)2 và (x+4)2 phải nhỏ nhất
=> (x-3)2=0(=>x-3=0=>x=3)
=> (x+4)2=0(=>x+4=0=>x=-4)
min (x-3)2+(x+4)2=0
\(\left(x-3\right)^2+\left(x+4\right)^2\)
\(=x^2-6x+9+x^2+8x+16\)
\(=2x^2+2x+25\)
\(=\left(\sqrt{2}x+\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{49}{2}\)
Vậy: Min là \(\dfrac{49}{2}\) khi \(x=\dfrac{-1}{2}\)
Đặt \(A=\left(x-3\right)^2+\left(x+4\right)^2\)
\(=x^2-6x+9+x^2+8x+16\)
\(=2x^2+2x+25\)
\(=2\left(x^2+x+\dfrac{25}{2}\right)\)
\(=2\left(x^2+x+\dfrac{1}{4}+\dfrac{49}{4}\right)\)
\(=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{49}{2}\)
Ta có: \(2\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Leftrightarrow2\left(x+\dfrac{1}{2}\right)^2+\dfrac{49}{2}\ge\dfrac{49}{2}\forall x\)
Dấu "=" xảy ra khi \(x+\dfrac{1}{2}=0\) hay \(x=-\dfrac{1}{2}\)
Vậy AMIN = \(\dfrac{49}{2}\) khi x = \(-\dfrac{1}{2}\).
. Cho x,y,z > 0. Tim min của A =\(4.\left(x^2+y^2+z^2\right)+\dfrac{441}{x+2y+4z}\)
điểm rơi xấu quá: x=\(\dfrac{\sqrt[3]{9}}{2}\); y=\(\sqrt[3]{9}\), z =\(2\sqrt[3]{9}\) (4x=2y=z)
Tim min A=x4 -x2 +2x+7
tim min y= (x^4+x^2+5)/(x^4+2x+1)
Tim Min B = \(\frac{x^4+x^2+5}{x^4+2x^2+1}\)
tim max hoac min cua x^2-2*x+y^2-4*y+16
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