\(4x^2+4x+2022=4x^2+4x+1+2021=\left(2x+1\right)^2+2021\ge2021\)

dấu "=" xảy ra \(< =>2x+1=0< =>x=\dfrac{-1}{2}\)

Đặt \(-6x^2+3x+3=0\)

\(\Leftrightarrow-6x^2+6x-3x+3=0\)

\(\Leftrightarrow-6x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Xe máy thứ nhất 1 giờ đi được 1/4 quảng đường

Xe máy thứ hai 1 giờ đi được 1/3 quảng đường

Sau 1,5 giờ 2 xe đi được:(1/4+1/3)x1,5=7/12x3/2=7/8(quảng đường)

quảng đường AB là:

15x8=120(km)

A=3x²+6x-1=3x²+6x+3-4=3(x+1)²-4

Do (x+1)²>0

=>3(x+1)²>0

=>A=3(x+1)²-4>-4

=>Min A=-4 <=>(x+1)²=0<=>x=-1

\(a,2x^2+y^2+6x-2xy+9=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+6x+9\right)=0\\ \Leftrightarrow\left(x-y\right)^2+\left(x+3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-3\end{matrix}\right.\Leftrightarrow x=y=-3\\ b,A=\left(x-2021\right)^2+\left(x+2022\right)^2=x^2-4042x+2021^2+x^2+4044x+2022^2\\ A=2x^2+2x+2021^2+2022^2\\ A=2\left(x^2+x+\dfrac{1}{4}\right)+2021^2+2022^2-\dfrac{1}{2}\\ A=2\left(x+\dfrac{1}{2}\right)^2+2021^2+2022^2-\dfrac{1}{2}\ge2021^2+2022^2-\dfrac{1}{2}\\ A_{max}=2021^2+2022^2-\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)\(c,P=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+16\\ P=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+16\\ P=\left(a^2+8a+11\right)^2-16+16=\left(a^2+8a+11\right)^2\left(Đpcm\right)\)

3x^2 + 6x - 1
= 3(x^2 + 2x - 1/3)
= 3(x^2 + 2x + 1 - 4/3)
= 3(x+1)^2 - 4 ≥ - 4
=> Cmin = - 4
Dấu "=" xảy ra khi x +  1 = 0<=> x = -1
Vậy Cmin = -4 khi x = -1

3x^2 + 6x - 1
= 3(x^2 + 2x - 1/3)
= 3(x^2 + 2x + 1 - 4/3)
= 3(x+1)^2 - 4 ≥ - 4
=> min = - 4
Dấu "=" xảy ra khi x +  1 = 0<=> x = -1
Vậy min = -4 khi x = -1

\(C=3x^2+6x-1\)

\(=3\left(x^2+2x\right)-1\)

\(=3\left(x^2+2x.1+1^2-1\right)-1\)

\(=3[\left(x+1\right)^2-1]-1\)

\(=3\left(x+1\right)^2-3-1\)

\(=3\left(x+1\right)^2-4\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\Rightarrow3\left(x+1\right)^2\ge0\Rightarrow C\ge-4\)

Dấu '' = '' xảy ra khi: \(\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)

Vậy \(MinC=-4\) chỉ khi \(x=-1\)

a) = 3(x²-2x+1) +1-3

GTNN = -2

B) tt