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b) Ta có : 4x - x2 + 1
= -(x2 - 4x - 1)
= -(x2 - 4x + 4 - 5)
= -(x2 - 4x + 4) + 5
= -(x - 2)2 + 5 \(\le5\forall x\) vì : \(-\left(x-2\right)^2\le0\forall x\)
Vậy GTLN của biểu thức là : 5 khi x = 2
Ta có : (x2 - 4xy + 4y2) + (10x - 20y) + (y2 - 2y + 1) + 27
= (x - 2y)2 + 10(x - 2y) + (y - 1)2
= (x - 2y)2 + 10(x - 2y) + 25 + (y - 1)2 + 2
= (x - 2y + 5)2 + (y - 1)2 + 2 \(\ge2\forall x\)
Vậy GTNN của biểu thức là 2
Khi \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
C = ( x2 - 4xy + 4y2 ) + 10.(x -2y) + ( y2 -2y + 1) + 27
= ( x-2y)2 + 2.5.(x-2y) + 25 + (y-1)2 + 2
= ( x-2y + 5 )2 + (y-1)2 + 2 \(\ge2\)vì \(\left(x-2y+5\right)^2\ge0\forall x,y\) và \(\left(y-1\right)^2\ge0\forall y\)
Dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
Vậy Min C = 2 \(\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
\(A=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)
\(A_{min}=3\) khi \(x=-2\)
\(B=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)
\(B_{min}=1\) khi \(x=10\)
\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
\(C_{min}=2\) khi \(\left(x;y\right)=\left(-3;1\right)\)
C = x2 - 4xy + 5y2 + 10x - 22y + 28
= (x2 - 4xy + 4y2) + (10x - 22y) + 25 + y2 + 3
= (x - 2y)2 + 10(x - 2y) + 25 + y2 + 3
= (x - 2y + 5)2 + y2 + 3 \(\ge\)3
Dấu " = " xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x-2y+5=0\\y=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=5\\y=0\end{cases}}\)
Vậy Min C = 3 \(\Leftrightarrow\)x = 5; y = 0
H=\(x^6-2x^3+x^2-2x+2\)
\(=x^6+2x^5+3x^4+2x^2-2x^5-4x^4-6x^3-4x^2-4x+x^4+2x^3+3x^2+2x+2\)
\(=x^2\left(x^4+2x^3+3x^2+2\right)-2x\left(x^4+2x^3+3x^2+2\right)+\left(x^4+2x^3+3x^2+2\right)\)
\(=\left(x^2-2x+1\right)\left(x^4+2x^3+3x^2+2\right)\)
\(=\left(x-1\right)^2\left(x^2+1\right)\left(x^2+2x+2\right)\)
\(=\left(x-1\right)^2\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]\text{≥}0\)
Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(x^2+1\right)\text{≥}1\\\left(x+1\right)^2+1\text{≥}1\end{matrix}\right.\)
⇒ MinH=0 ⇔ \(x=1\)
B=[(x - 2)(x - 5)](x2– 7x - 10)
= (x2- 7x + 10)(x2 - 7x - 10)
= (x2 - 7x)2- 102
= (x2 - 7x)2 - 100
=>(x2-7x)2\(\ge\) 100
GTNN = -100 \(\Rightarrow\) x2 - 7x = 0 \(\Leftrightarrow\) x(x-7) = 0 \(\Leftrightarrow\) x = 0 hoặc x = 7
B = x2 - 4xy + 5y2 + 10x - 22y + 28
= x2 - 4xy + 4y2+ y2+ 10(x-2y) + 28
= (x - 2y)2+ 10(x-2y) + 25 + y2- 2y+ 1 + 2
= (x-2y + 5)2 + (y-1)2 + 2\(\ge\) 2
GTNN B = 2, khi y=1, x=-3
\(A=x-x^2=-\left(x^2-2\times x\times\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(\left(x-\frac{1}{2}\right)^2\ge0\)
\(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
\(-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\le\frac{1}{4}\)
Vậy Max A = \(\frac{1}{4}\) khi x = \(\frac{1}{2}\)
***
\(B=5-8x-x^2=-\left(x^2+2\times x\times4+4^2-4^2-5\right)=-\left[\left(x+4\right)^2-21\right]\)
\(\left(x+4\right)^2\ge0\)
\(\left(x+4\right)^2-21\ge-21\)
\(-\left[\left(x+4\right)^2-21\right]\le21\)
Vậy Max B = 21 khi x = - 4
***
\(C=5-x^2+2x-4y^2-4y=-\left(x^2-2\times x\times1+1^2-1^2+\left(2y\right)^2-2\times2y\times1+1^2-1^2-5\right)=-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\)
\(\left(x-1\right)^2\ge0\)
\(\left(2y-1\right)^2\ge0\)
\(\left(x-1\right)^2+\left(2y-1\right)^2-7\ge-7\)
\(-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\le7\)
Vậy Max C = 7 khi x = 1 và y = \(\frac{1}{2}\)