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a: Ta có: \(A=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x-10\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=10
\(VT=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\left(x-y\right)^2\left(x+y\right)^2=VP\)
VT\(=\left(x^2+y^2-2xy\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)^2\cdot\left(x+y\right)^2\)
\(A=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)
\(A_{min}=3\) 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 \(\left(x;y\right)=\left(-3;1\right)\)
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
Bài 1:
a) Ta có: \(A=-x^2-4x-2\)
\(=-\left(x^2+4x+2\right)\)
\(=-\left(x^2+4x+4-2\right)\)
\(=-\left(x+2\right)^2+2\le2\forall x\)
Dấu '=' xảy ra khi x=-2
b) Ta có: \(B=-2x^2-3x+5\)
\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)
\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)
\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)
c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)
\(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9\le9\forall x\)
Dấu '=' xảy ra khi x=-1
Bài 2:
a) Ta có: \(=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)
b) Ta có: \(B=9x^2-6xy+2y^2+1\)
\(=9x^2-6xy+y^2+y^2+1\)
\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)
c) Ta có: \(E=x^2-2x+y^2-4y+6\)
\(=x^2-2x+1+y^2-4y+4+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
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 ⇔ \(x=1\)
\(A=x^2-20x+101=\left(x-10\right)^2+1\ge1\)
\(minA=1\Leftrightarrow x=10\)
\(B=2x^2+40x-1=2\left(x+10\right)^2-201\ge-201\)
\(minB=-201\Leftrightarrow x=-10\)
\(C=x^2-4xy+5y^2-2y+28=\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+27=\left(x-2y\right)^2+\left(y-1\right)^2+27\ge27\)
\(minC=27\Leftrightarrow\)\(\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
\(D=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x+10\right)=\left(x^2-7x\right)^2-100\ge-100\)
\(minD=100\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
a: Ta có: \(A=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x-10\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=10
b: ta có: \(B=2x^2+40x-1\)
\(=2\left(x^2+20x-\dfrac{1}{2}\right)\)
\(=2\left(x^2+20x+100-\dfrac{201}{2}\right)\)
\(=2\left(x+10\right)^2-201\ge-201\forall x\)
Dấu '=' xảy ra khi x=-10