B=y^2-y+1

=y^2-2*y*1/2+1/4+3/4

=(y-1/2)^2+3/4>=3/4

Dấu = xảy ra khi y=1/2

E=-x^2+x+2

=-(x^2-x-2)

=-(x^2-x+1/4-9/4)

=-(x-1/2)^2+9/4<=9/4

Dấu = xảy ra khi x=1/2

\(-2x+4\sqrt{x}+1\)

\(=-2\left(x-2\sqrt{x}+1\right)+3\)

\(=-2\left(\sqrt{x}-1\right)^2+3\le3\left(\forall x\ge0\right)\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

 ĐKXĐ :\(x\ge0\)

\(x-4\sqrt{x}+5\)

\(=x-4\sqrt{x}+4+1\)

\(=\left(\sqrt{x}-2\right)^2+1\ge1\forall x\ge0\)

Dấu"=" xả ra <=> \(\left(\sqrt{x}-2\right)^2=0\)

                    \(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

B = 4x² + 8x 

= 4( x² + 2x + 1 ) - 4

= 4( x + 1 )² - 4

4( x + 1 )² ≥ 0 ∀ x => 4( x + 1 )² - 4 ≥ -4

Đẳng thức xảy ra <=> x + 1 = 0 => x = -1

=> MinB = -4 <=> x = -1

C = -2x² + 8x - 15

= -2( x² - 4x + 4 ) - 7

= -2( x - 2 )² - 7

-2( x - 2 )² ≤ 0 ∀ x => -2( x - 2 )² - 7 ≤ -7

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MaxC = -7 <=> x = 2

A=2(x2+2.x.4+16)−49≥−49A=2(x2+2.x.4+16)−49≥−49.Dấu "=" xảy ra khi x=−4x=−4

tk nhé

Ta có: A = 2x² + 4x + 5 = 2(x² + 2x + 1) + 3 = 2(x + 1)² + 3 \(\ge\)3 \(\forall\)x

Dấu "=" xảy ra <=> x + 1 = 0 <=> x = -1

Vậy MinA = 3 <=> x = -1

\(2x^2+4x+5\)

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

\(=2\left(x^2+2x+1+\frac{3}{2}\right)\)

\(=2\left[\left(x+1\right)^2+\frac{3}{2}\right]\)

\(=2\left(x+1\right)^2+3\ge3\)

Dấu '' = '' xảy ra khi 

\(\Leftrightarrow2\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy............................

P/s : sai thì thôi nha