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

\(=\left(x+2-y\right)\left(x+2+y\right)\)

\(B=-2x^2+10x-8=-2x^2+10x-\frac{25}{2}+\frac{9}{2}\)

\(=-\left(2x^2-10x+\frac{25}{2}\right)+\frac{9}{2}\)

\(=-2\left(x^2-5x+\frac{25}{4}\right)+\frac{9}{2}\)

\(=-2\left[x^2-2.\frac{5}{2}x+\left(\frac{5}{2}\right)^2\right]+\frac{9}{2}\)

\(=-2\left(x-\frac{5}{2}\right)^2+\frac{9}{2}\)

Vì \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)\(\Rightarrow-2\left(x-\frac{5}{2}\right)^2\le0\forall x\)

\(\Rightarrow-2\left(x-\frac{5}{2}\right)^2+\frac{9}{2}\le\frac{9}{2}\forall x\)

hay \(B\le\frac{9}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\)\(\Leftrightarrow x=\frac{5}{2}\)

Vậy \(maxB=\frac{9}{2}\)\(\Leftrightarrow x=\frac{5}{2}\)

Ta có: 

\(B=-2x^2+10x-8\)

\(B=-2\left(x^2-5x+\frac{25}{4}\right)+\frac{9}{2}\)

\(B=-2\left(x-\frac{5}{2}\right)^2+\frac{9}{2}\le\frac{9}{2}\left(\forall x\right)\)

Dấu "=" xảy ra khi: x = 5/2

Vậy Max(B) = 9/2 khi x = 5/2

B=-2(x^2-5x+25/4)-7/4=-2(x-5/2)^2-7/4

Tự làm tiếp

Ta có: \(2x^2-10x+14\)

\(=2\left(x^2-5x+\frac{25}{4}\right)+\frac{3}{2}\)

\(=2\left(x-\frac{5}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}>0\left(\forall x\right)\)

\(\Rightarrow2x^2-10x+14>0\)

GTLN chứ ? 

B = -2x² + 10x - 8

= -2( x² - 5/2x + 25/4 ) + 9/2

= -2( x - 5/2 )² + 9/2 ≤ 9/2 ∀ x

Dấu "=" xảy ra khi x = 5/2

=> MaxB = 9/2 <=> x = 5/2

Đề phải là tìm GTLN nhé

Ta có: 

\(B=-2x^2+10x-8\)

\(B=-2\left(x^2-5x+\frac{25}{4}\right)+\frac{9}{2}\)

\(B=-2\left(x-\frac{5}{2}\right)^2+\frac{9}{2}\le\frac{9}{2}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(-2\left(x-\frac{5}{2}\right)^2=0\Rightarrow x=\frac{5}{2}\)

Vậy Max(B) = 9/2 khi x = 5/2

Ta có

   \(2x^2-10x+14=\left(2x^2-10x\right)+14\)

                                       \(=2\left(x^2-5x\right)+14\)

                                        \(=2\left(x^2-2.\frac{5}{2}.x+\frac{25}{4}-\frac{25}{4}\right)+14\)

                                        \(=2\left(x-\frac{5}{2}\right)^2+14-\frac{50}{4}\)

                                          \(=2\left(x-\frac{5}{2}\right)^2+\frac{3}{2}\)

  Vì \(2\left(x-\frac{5}{2}\right)^2\ge0\forall x\)

       \(\frac{3}{2}>0\)

  Nên \(2\left(x-\frac{5}{2}\right)^2+\frac{3}{2}>0\forall x\left(đpcm\right)\)

Ta có : 2x² - 10x + 14

= 2( x² - 5x + 25/4 ) + 3/2

= 2( x - 5/2 )² + 3/2 ≥ 3/2 > 0 ∀ x

=> đpcm

sửa các đa thức theo thứ tự là x^2 - 6x + 8 ; x - 2