a,\(x^2+4x+7=x^2+4x+4+3=\left(x+2\right)^2+3\ge3\)
Dấu = xảy ra \(< =>x+2=0< =>x=-2\)
Vậy \(A_{min}=3\)khi \(x=-2\)
b,\(4x^2+4x+6=\left(2x\right)^2+4x+1+5=\left(2x+1\right)^2+5\ge5\)
Dấu = xảy ra \(< =>2x+1=0< =>x=-\frac{1}{2}\)
Vậy \(B_{min}=5\)khi \(x=-\frac{1}{2}\)
c,\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra \(< =>x+\frac{1}{2}=0< =>x=-\frac{1}{2}\)
Vậy \(C_{min}=\frac{3}{4}\)khi \(x=-\frac{1}{2}\)
d,\(2x^2-6x=2\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
Dấu = xảy ra \(< =>x-\frac{3}{2}=0< =>x=\frac{3}{2}\)
Vậy \(D_{min}=-\frac{9}{2}\)khi \(x=\frac{3}{2}\)
A=x2+4x+4+3=(x+2)2+3
1. \(A=x^2+4x+7=\left(x+2\right)^2+3\)
Vì \(\left(x+2\right)^2\ge0\forall x\)\(\Rightarrow\left(x+2\right)^2+3\ge3\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x=-2\)
Vậy minA = 3 <=> x = - 2
2. \(B=4x^2+4x+6=4\left(x+\frac{1}{2}\right)^2+5\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow4\left(x+\frac{1}{2}\right)^2+5\ge5\)
Dấu "=" xảy ra \(\Leftrightarrow4\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy minB = 5 <=> x = - 1/2
3. \(C=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy minC = 3/4 <=> x = - 1/2
4. \(D=2x^2-6x=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow2\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Vậy minD = - 9/2 <=> x = 3/2
A = x2 + 4x + 7 = ( x2 + 4x + 4 ) + 3 = ( x + 2 )2 + 3
( x + 2 )2 ≥ 0 ∀ x => ( x + 2 )2 + 3 ≥ 3
Đẳng thức xảy ra <=> x + 2 = 0 => x = -2
=> MinA = 3 <=> x = -2
B = 4x2 + 4x + 6 = ( 4x2 + 4x + 1 ) + 5 = ( 2x + 1 )2 + 5
( 2x + 1 )2 ≥ 0 ∀ x => ( 2x + 1 )2 + 5 ≥ 5
Đẳng thức xảy ra <=> 2x + 1 = 0 => x = -1/2
=> MinB = 5 <=> x = -1/2
C = x2 + x + 1 = ( x2 + x + 1/4 ) + 3/4 = ( x + 1/2 )2 + 3/4
( x + 1/2 )2 ≥ 0 ∀ x => ( x + 1/2 )2 + 3/4 ≥ 3/4
Đẳng thức xảy ra <=> x + 1/2 = 0 => x = -1/2
=> MinC = 3/4 <=> x = -1/2
D = 2x2 - 6x = 2( x2 - 3x + 9/4 ) - 9/2 = 2( x - 3/2 )2 - 9/2
2( x - 3/2 )2 ≥ 0 ∀ x => 2( x - 3/2 )2 - 9/2 ≥ -9/2
Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2
=> MinD = -9/2 <=> x = 3/2
