a. M = 9 - 6x + x²

= ( 3 - x )²\(\ge\)0\(\forall\)x

=> Đpcm

b. B = 4x² + 4x + 2007

= 4x² + 4x + 1 + 2006

= 4 ( x +\(\frac{1}{2}\))² + 2006\(\ge\)2006\(\forall\)x

=> Đpcm

M = 9 - 6x + x²

= ( 3 - x )² ≥ 0 ∀ x ( đpcm )

B = 4x² + 4x + 2007

= ( 4x² + 4x + 1 ) + 2006

= ( 2x + 1 )² + 2006 ≥ 2006 > 0 ∀ x ( đpcm )

\(B=-2x^2+4x-5\)

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

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

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

\(=-2\left[\left(x-1\right)^2\right]-3\le3< 0\forall x\)

\(B=-2x^2+4x-5\)

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

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

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

\(B=-2\left(x-1\right)^2-3\)

Mà \(\hept{\begin{cases}-2\left(x-1\right)^2\le0\forall x\\-3< 0\end{cases}\Rightarrow B< 0\forall x}\)

\(1-\left(\frac{a+x}{ax-x^2}+\frac{2a+3x}{x^2-a^2}\right):\frac{a^4-4x^4}{a^4x-a^2x^3}\)

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

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

\(=1-\frac{a^2+2\cdot ax+x^2-2ax-3x^2}{\left(a^2-2x^2\right)\left(a^2+2x^2\right)}\cdot a^2=1-a^2\cdot\frac{a^2-2x^2}{\left(a^2-2x^2\right)\left(a^2+2x^2\right)}\)

\(=1-\frac{a^2}{a^2+2x^2}=\frac{a^2+2x^2-a^2}{a^2+2x^2}=\frac{2x^2}{a^2+2x^2}\)

a: \(A=x^3-27-x^3+3x^2-3x+1-4\left(x^2-4\right)-x\)

\(=3x^2-4x-26-4x^2+16\)

\(=-x^2-4x-10\)