\(A=2x^2+6x\)
\(A=2\left(x^2+3x\right)\)
\(A=2\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
\(A=2\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{9}{4}.2\)
\(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge\dfrac{9}{2}\)
Dấu "=" xảy ra khi :
\(x=-\dfrac{3}{2}\)
B đã sửa đề vì theo đề của you thì ko có tổng nào = nhau
\(B=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(B=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(B=\left[x\left(x+4\right)+1\left(x+4\right)\right]\left[x\left(x+3\right)+2\left(x+3\right)\right]\)
\(B=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)\)
\(B=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
\(B=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)\)
\(B=\left(x^2+5x+5\right)^2-1\ge-1\)
Dấu "=" xảy ra khi:
\(x^2+5x+5=0\)
Cạn....
\(A=2x^2+6x+\dfrac{9}{2}-\dfrac{9}{2}\\ =2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
Với mọi x thì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\\ \Rightarrow2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Hay \(A\ge-\dfrac{9}{2}\)
Để \(A=-\dfrac{9}{2}\) thì \(\left(x+\dfrac{3}{2}\right)^2=0\)
=>\(x+\dfrac{3}{2}=0\)
=>\(x=-\dfrac{3}{2}\)
Vậy...
A = 2x2 + 6x
= 2 x2 + 3x)
= 2(x2 + 2.1,5.x + 2,25 - 2,25)
= 2(x2 + 2.1,5.x + 2,25) - 2.2,25
= 2(x+1,5)2 - 4,5
Xét 2(x+1,5)2 \(\ge\) 0
=> 2(x+1,5)2 - 4,5 \(\le\) -4,5
Vậy MinA = -4,5 khi x = -1,5
B sai đề nhé! Mình sửa lại:
B=(x-1)(x+2)(x+3)(x+6)
= (x-1)(x+6)(x+2)(x+3)
= (x2+5x-6)(x2+5x+6)
Đặt x2+5x=a
=> (a-6)(a+6)
= a2 - 36\(\ge\) -36
=> MinB = -36 khi x = 0 hoặc x=5
Ta có:A=\(2x^2+6x=2\left(x^2+3x\right)\)
\(=2.\left[x^2+2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\)
\(=2\left[\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\)
\(=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
Do \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) với \(\forall x\) (dấu "=" xảy ra \(\Leftrightarrow x+\dfrac{3}{2}=0\Rightarrow x=-\dfrac{3}{2}\))
\(\Rightarrow2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\) hay A\(\ge-\dfrac{9}{2}\) (dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{3}{2}\))
Vậy Min A=\(-\dfrac{9}{2}\) tại x=\(-\dfrac{3}{2}\)
