2.E = 4x^2 -  12x

= ( 4x^2 - 12x + 9 ) -9

=(2x-3)^2 - 9 >= -9 

<=> E >= -18 

Dấu "=" xảy ra <=> 2x-3 = 0 <=> x=3/2

Vậy GTNN của E là E = -18 <=> x =3/2

Ta có : E = 2x² - 6x 

=> E = 2(x² - 6x + 9 - 9)

=> E = 2(x² - 6x + 9) - 18

=> E = 2(x - 3)² - 18

Mà ;  2(x - 3)² \(\ge0\forall x\)

Nên: E = 2(x - 3)² - 18 \(\ge-18\forall x\)

Vậy E_min = -18 khi x = 3

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

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

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

Đặt:

\(A=2x^2-6x\)

\(A=2x^2-6x+\dfrac{9}{2}-\dfrac{9}{2}\)

\(A=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\)

\(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

Vì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) nên \(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" xảy ra khi:

\(x=-\dfrac{3}{2}\)

\(2x^2-6x\)

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

=\(2\left[x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3^{ }}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\right]\)

\(=2\left[\left(x-\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\right]\)

=\(2\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\ge2\left(0-\dfrac{9}{4}\right)\ge0\)

Vậy GTNN của biểu thức là\(\dfrac{-9}{2}\) xẩy ra khi \(x=\dfrac{3}{2}\)

Nguồn: OLM

Bạn học tốt nhé!

\(2x^2-6x\\ =2x^2-6x+\dfrac{9}{2}-\dfrac{9}{2}\\ =2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ =2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{9}{2}\\ =2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

\(\text{Ta có: }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)

Dấu \("="\) xảy ra khi:

\(2\left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow \left(x-\dfrac{3}{2}\right)^2=0\\\Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\)

Vậy \(GTNN\) của biểu thức là \(-\dfrac{9}{2}\) khi \(x=\dfrac{3}{2}\)

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

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

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

\(5x=5\)

\(x=1\)

b) \(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\)

\(4x^4-6x^3-4x^4+6x^2-2x^2=0\)

\(-2x^2=0\)

\(x^2=0\)

\(x=0\)

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

\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

\(=x^5+x+1\)