Tìm Min
F=2x^2+2y^2-x-y+11
Tìm Min
F=2x^2+2y^2-x-y+11
\(F=2x^2+2y^2-x-y+11\)
\(F=2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+2\left(y^2-\frac{1}{2}y+\frac{1}{16}\right)+\frac{87}{8}\)
\(F=2\left(x-\frac{1}{4}\right)^2+2\left(y-\frac{1}{4}\right)^2+\frac{87}{8}\ge\frac{87}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}2\left(x-\frac{1}{4}\right)^2=0\\2\left(y-\frac{1}{4}\right)^2=0\end{cases}\Leftrightarrow x=y=\frac{1}{4}}\)
Vậy GTNN của \(F\) là \(\frac{87}{8}\) khi \(x=y=\frac{1}{4}\)
Chúc bạn học tốt ~
\(Tìm Min : B=2x²-4x-8 C=x²-2xy+2y²+2x-10y+17 D=x²-xy+y²-2x-2y E=(x²+x-6)(x²+x+2) F=(x+1)(x+2)(x+3)(x+4) Tìm Max G= 4x-x2 H=25-x-5x2 \)
F=(11 2x−2y 2(x−y)−1):(2x−2y−(4x2−8xy 4y22x−2y 1))F=(11 2x−2y 2(x−y)−1):(2x−2y−(4x2−8xy 4y22x−2y 1))F=\left(\dfrac{1}{1 2x-2y} 2\left(x-y\right)-1\right):\left(2x-2y-\left(\dfrac{4x^2-8xy 4y^2}{2x-2y 1}\right)\right) Cm giá trị của F là một số chẵn vs mọi x,
Tìm Min :
B=2x²-4x-8
C=x²-2xy+2y²+2x-10y+17
D=x²-xy+y²-2x-2y
E=(x²+x-6)(x²+x+2)
F=(x+1)(x+2)(x+3)(x+4)
Tìm Max
G= 4x-x2
H=25-x-5x2
\(B=2x^2-4x-8=2\left(x^2-2x-4\right)\)
\(=2\left(x^2-2x+1-5\right)\)
\(=2\left[\left(x-1\right)^2-5\right]\)
\(=2\left(x-1\right)^2-10\ge-10\)
Vậy \(B_{min}=-10\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(F=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
Đặt \(x^2+5x+4=t\)
\(\RightarrowĐT=t\left(t+2\right)=t^2+2t+1-1\)
\(=\left(t+1\right)^2-1\ge-1\)
hay \(\left(x^2+5x+5\right)^2-1\ge-1\)
Vậy \(F_{min}=-1\Leftrightarrow x^2+5x+5=0\)
\(\Leftrightarrow x^2+5x+\frac{25}{4}-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}}-\frac{5}{2}\\x=-\sqrt{\frac{5}{4}}-\frac{5}{2}\end{cases}}\)
\(G=4x-x^2=-\left(x^2-4x+4-4\right)\)
\(=-\left[\left(x-2\right)^2-4\right]=-\left(x-2\right)^2+4\le4\)
Vậy \(G_{max}=4\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(H=25-x-5x^2=-5\left(x^2+\frac{x}{5}-5\right)\)
\(=-5\left(x^2+2x.\frac{1}{10}+\frac{1}{100}-\frac{501}{100}\right)\)
\(=-5\left[\left(x+\frac{1}{10}\right)^2-\frac{501}{100}\right]\)
\(=-5\left(x+\frac{1}{10}\right)^2+\frac{101}{20}\le\frac{101}{2}\)
Vậy \(H_{max}=\frac{101}{2}\Leftrightarrow x+\frac{1}{10}=0\Leftrightarrow x=-\frac{1}{10}\)
f(x)=(2x-3)^2+(x+4)^2-(3x^2+5x-2) tìm GTNN
F=2x^2+3y^2-8x+24y-7 tìm GTNN
F=-5x^2-4y^2+20x-32y+9 tìm GTLN
F=x^2+y^2-x+y-3 tìm GTNN
F=F=5x^2+y^2-4xy-6x+20 tìm GTNN
F=-13x^2-4y^2+12xy+20x+37
F=5x^2+9y^2-12xy+24x-48y+100
Cho x+y=5 Cho A= x^3+y^3-8(x^2+y^2)+xy+2 tính GTLN của A
Cho x+y+2=0 Tìm min của B=2(x^3+y^3)-15xy+7
Cho x+y+2=0 tìm min của C=x^4+y^4-(x^3+y^3)+2x^2y^2+2xy(x^2+y^2)+13xy
tìm min 2x^2+y^2-xy+x-2y
\(2x^2+y^2-xy+x-2y=\dfrac{1}{4}x^2-x\left(y-1\right)+\left(y-1\right)^2+\dfrac{7}{4}x^2-1=\left(\dfrac{1}{2}x-y+1\right)^2+\dfrac{7}{4}x^2-1\ge-1\)
\(min=-1\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)
tìm min
A = x2+5y2-4xy-y+7
B = 5x2+y2+4xy-2x-4y-11
C = x2+2y2-2xy+4x-2y+3
giúp mìn với
x^2 +2y^2 +2xy +2x+2y -3 = O
Tìm max min của Q = x+y
Tìm Min A = \(\sqrt{x^2+2y^2-6x+4y+11}+\sqrt{x^2+3y^2+2x+6y+4}\)
Ta có:
\(A=\sqrt{\left(x-3\right)^2+2\left(y+1\right)^2}+\sqrt{\left(x+1\right)^2+3\left(y+1\right)^2}\)
Áp dụng bđt Minkowski, ta có:
\(\Rightarrow A=\sqrt{\left(x-3\right)^2+2\left(y+1\right)^2}+\sqrt{\left(x+1\right)^2+3\left(y+1\right)^2}\)
\(A=\sqrt{\left(3-x\right)^2+2\left(y+1\right)^2}+\sqrt{\left(x+1\right)^2+3\left(y+1\right)^2}\)\(\ge\sqrt{\left(3-x+x+1\right)^2+\left(\sqrt{2}+\sqrt{3}\right)^2\left(y+1\right)^2}\)
\(A=\sqrt{4^2+\left(\sqrt{2}+\sqrt{3}\right)^2\left(y+1\right)^2}\ge\sqrt{4^2}=4\)
\(\Rightarrow A\ge4.Đ\text{TXR}\Leftrightarrow\orbr{\begin{cases}x=1;y=-1\\x=3;y=-1\end{cases}}\)
Dấu "=" xảy ra khi (x; y) = (3; -1)