A = x² + x + 1 = ( x² + x + 1/4 ) + 3/4 = ( x + 1/2 )² + 3/4 ≥ 3/4 ∀ x

Dấu "=" xảy ra khi x = -1/2

=> MinA = 3/4 <=> x = -1/2

B = -x² - 4x + 12 = -( x² + 4x + 4 ) + 16 = -( x + 2 )² + 16 ≤ 16 ∀ x

Dấu "=" xảy ra khi x = -2

=> MaxB = 16 <=> x = -2

C = \(\frac{5}{x^2+6}\)

Ta có : x² + 6 ≥ 6 ∀ x

<=> \(\frac{1}{x^2+6}\le\frac{1}{6}\forall x\)

<=> \(\frac{5}{x^2+6}\le\frac{5}{6}\forall x\)

Dấu "=" xảy ra khi x = 0

=> MaxC = 5/6 <=> x = 0

a, \(B=\left(\frac{2x+1}{2x-1}+\frac{4}{1-4x^2}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)

\(=\left(\frac{2x+1}{2x-1}+\frac{4}{\left(1-2x\right)\left(2x+1\right)}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)

\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{4}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right):\frac{x^2+2}{2x+1}\)

\(=\left(\frac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\right):\frac{x^2+2}{2x+1}\)

\(=\frac{8x-4}{\left(2x-1\right)\left(2x+1\right)}.\frac{2x+1}{x^2+2}=\frac{8x-4}{\left(2x-1\right)\left(x^2+2\right)}\)

b, Thay x = -1 ta được : \(\frac{9\left(-1\right)-4}{\left[2\left(-1\right)-1\right]\left[\left(-1\right)^2+2\right]}=-\frac{13}{-9}=\frac{13}{9}\)

1) Ta có: 3x - x² = -(x² - 3x + 9/4) + 9/4 = -(x - 3/2)² + 9/4

Ta luôn có: -(x - 3/2)² \(\le\)0 \(\forall\)x

=> -(x - 3/2)² + 9/4 \(\le\)9/4 \(\forall\)x

Dấu "=" xảy ra <=> x - 3/2 = 0 <=> x = 3/2

Vậy Max của 3x - x² là 9/4 tại x = 3/2

2) Ta có : -(x² + y²) + x + 3y+  10 = -x² - y² + x + 3y + 10 = -(x² - x + 1/4) - (y² -3y + 9/4) + 25/2 = -(x - 1/2)² - (y - 3/2)² + 25/2

Ta luôn có: -(x - 1/2)² \(\le\)0 \(\forall\)x

           -(y - 3/2)² \(\le\)0 \(\forall\)y

=> -(x - 1/2)² - (y - 3/2)² + 25/2 \(\le\)25/2 \(\forall\)x;y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{3}{2}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{2}\end{cases}}\)

Vậy ...