a) \(A=\left(x-3\right)\left(x+5\right)+20\)
\(\Leftrightarrow A=x^2+5x-3x-15+20\)
\(\Leftrightarrow A=x^2+2x+5\)
\(\Leftrightarrow A=x^2+2x+1+4\)
\(\Leftrightarrow A=\left(x+1\right)^2+4\ge4\)
GTNN của A = 4
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy ..........................
c, đề : \(C=x^2+2x+1\) đước ko chị ?
\(A=\left(x-3\right)\left(x+5\right)+20\)
\(=x^2+5x-3x-15+20\)
\(=x^2+2x+5=x^2+2x+1+4\)
\(=\left(x+1\right)^2+4\ge4\)
Dấu = xảy ra \(< =>x=-1\)
Vậy \(A_{mịn}=4\)khi \(x=-1\)
b) \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(\Leftrightarrow B=\left(x^2+2x-x-2\right)\left(x^2+6x+3x+18\right)\)
\(\Leftrightarrow B=\left(x^2+x-2\right)\left(x^2+9x+18\right)\)
\(\Leftrightarrow B=x^4+9x^3+18x^2+x^3+9x^2+18x-2x^2-18x-36\)
\(\Leftrightarrow B=x^4+10x^3+25x^2-36\)
\(\Leftrightarrow B=x^2\left(x^2+10x+25\right)-36\)
\(\Leftrightarrow B=x^2\left(x+5\right)^2-36\ge-36\)
GTNN của B = -36
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}}\)
Vậy .............
c) \(C=x^2+x+1\)
\(\Leftrightarrow C=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(\Leftrightarrow C=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
GTNN của \(C=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy .................
\(C=x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra khi và chỉ khi \(x=-\frac{1}{2}\)
Vậy \(C_{min}=-\frac{1}{2}\)đạt được khi \(x=-\frac{1}{2}\)
Bài giải
\(a,\text{ }A=\left(x-3\right)\left(x+5\right)+20\)
\(A=x^2-3x+5x-15+20\)
\(A=x^2+2x+5=x^2+2x+1+4\)
\(A=\left(x+1\right)^2+4\ge4\)
Dấu " = " xảy ra khi :
\(\left(x+1\right)^2=0\text{ }\Rightarrow\text{ }x+1=0\text{ }\Rightarrow\text{ }x=-1\)
\(\Rightarrow\text{ }Min_A=4\text{ khi }x=-1\)
\(b,\text{ }B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(B=\left(x^2-x+2x-2\right)\left(x^2+3x+6x+18\right)\)
\(B=\left(x^2+x-2\right)\left(x^2+9x+18\right)\)
\(B=x^4+x^3-2x^2+9x^3+9x^2-18x+18x^2+18x-36\)
\(B=x^4+10x^3+25x^2-36\)
\(B=x^2\left(x^2+10x+25\right)-36=x^2\left(x+5\right)^2-36\ge-36\)
Dấu " = " xảy ra khi \(x^2\left(x+5\right)^2=0\)\(\Rightarrow\orbr{\begin{cases}x^2=0\\\left(x+5\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy \(Min_B=-36\text{ khi }x=0\text{ hoặc }x=-5\)
\(c,\text{ }C=x^2+x+1\)
\(=x^2+2\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu " = " xảy ra khi \(\left(x+\frac{1}{2}\right)^2=0\text{ }\Rightarrow\text{ }x+\frac{1}{2}=0\text{ }\Rightarrow\text{ }x=-\frac{1}{2}\)
Vậy \(Min_C=\frac{3}{4}\text{ khi }x=-\frac{1}{2}\)
A = ( x - 3 )( x + 5 ) + 20
A = x2 + 2x - 15 + 20
A = x2 + 2x + 1 + 4
A = ( x + 1 )2 + 4
\(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2+4\ge4\)
Dấu " = " xảy ra <=> x + 1 = 0 => x = -1
Vậy MinA = 4 , đạt được khi x = -1
B = ( x - 1 )( x + 2 )( x + 3 )( x + 6 )
B = [( x - 1 )( x + 6 )][( x + 2 )( x + 3 )]
B =( x2 + 5x - 6 )( x2 + 5x + 6 )
B = ( x2 + 5x )2 - 36
\(\left(x^2+5x\right)^2\ge0\forall x\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)
Dấu " = " xảy ra <=> x2 + 5x = 0
<=> x( x + 5 ) = 0
<=> x = 0 hoặc x = -5
Vậy MinB = -36, đạt được khi x = 0 hoặc x = -5
C = x2 + x + 1
C = x2 + 2.1/2.x + 1/4 + 3/4
C = ( x + 1/2 )2 + 3/4
\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu " = " xảy ra <=> x + 1/2 = 0 => x = -1/2
Vậy MinC = 3/4 , đạt được khi x = -1/2
