a, x^2-2x+1+4=(x-1)^2+4>=4. dấu = xảy ra khi x=1
b,dưa 2 ra làm tt
c, đưa dấu - ra
d nhân ra là đc
\(x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\left(x-1\right)^2+4\ge4\forall x\)
\(x^2-2x+5=4\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy Min \(x^2-2x+5=4\Leftrightarrow x=1\)
Câu b trình bày tương tự, mình chỉ gợi ý
\(2x^2+6x-5=2\left(x^2+2.x.1,5+1,5^2\right)-9,5=2\left(x+1,5\right)^2-9,5\)
a) x^2 - 2x + 5
= x^2 - 2x + 1 + 4
= (x - 1)^2 + 4 ≥ 4
=> Amin = 4. Dấu " = " xảy ra khi x - 1 = 0 <=> x = 1
Vậy Amin = 4 khi x = 1
b) 2x^2 + 6x - 5
= 2(x^2 + 3x - 5/2)
= 2(x^2 + 2x3/2 + 9/4 -19/4)
= 2(x - 3/2)^2 - 19/2 ≥ -19/2
=> Bmin = -19/2. Dấu "=" xảy ra khi x - 3/2 = 0 <=> x = 3/2
Vậy Bmin = -19/2 khi x = 3/2
c) -x^2 - 4x - 2
<=>-(x+2)^2 +2 <=2
=>max =2 <=>x=-2
d) (2 - x)(x + 4)
=-x^2 -2x +8=-(x+1)^2 +9 <=9
=> max =9 <=>x=-1
\(-x^2-4x-2\)
\(=-\left(x^2+2.x.2+2^2\right)+2\)
\(=-\left(x+2\right)^2+2\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x+2\right)^2\le0\forall x\)
\(\Leftrightarrow-\left(x+2\right)^2+2\le2\forall x\)
\(-x^2-4x-2=2\Leftrightarrow-\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy Max \(-x^2-4x-2=2\Leftrightarrow x=-2\)
d) \(\left(2-x\right)\left(x+4\right)=-x^2-2x+8=-\left(x^2+2x+1\right)+7=-\left(x+1\right)^2+7\le7\)
