a) \(M(x) = A(x) + B(x) \\= 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 - 5{x^2} + 7{x^3} + 5x + 4 - 4{x^4} \\=(4x^4-4x^4)+(-7x^3+7x^3)+(6x^2-5x^2)+(-5x+5x)+(-6+4)\\= {x^2} - 2.\)

b) \(A(x) = B(x) + C(x) \Rightarrow C(x) = A(x) - B(x)\)

\(\begin{array}{l}C(x) = A(x) - B(x)\\ = 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 - ( - 5{x^2} + 7{x^3} + 5x + 4 - 4{x^4})\\ = 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 + 5{x^2} - 7{x^3} - 5x - 4 + 4{x^4}\\ =(4x^4+4x^4)+(-7x^3-7x^3)+(6x^2+5x^2)+(-5x-5x)+(-6-4)\\= 8{x^4} - 14{x^3} + 11{x^2} - 10x - 10\end{array}\) 

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

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

\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)

\(=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)

TH1: x>=8

=>x-4>=4

=>\(\sqrt{x-4}\ge2\)

=>\(\sqrt{x-4}-2\ge0\)

\(M=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)

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

TH2: 4<=x<8

=>\(\sqrt{x-4}-2<0\)

\(M=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)

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

=4

\(\)Ta có: \(x+y+z+\sqrt{xyz}=4\)

=>\(y + z = 4 - x - \sqrt{xyz}\)

Ta có: \((4-y)(4-z)=16-4(y+z)+yz\)

=>\((4-y)(4-z) = 16 - 4(4 - x - \sqrt{xyz}) + yz\)

\(= 16 - 16 + 4x + 4\sqrt{xyz} + yz\)

\(= 4x + 4\sqrt{xyz} + yz\)

=>\(x(4-y)(4-z) = x(4x + 4\sqrt{xyz} + yz)\)

\(=4x^2+4x\sqrt{xyz}+xyz=\left(2x+\sqrt{xyz}\right)^2\)

=>\(\sqrt{x(4-y)(4-z)} = \sqrt{(2x + \sqrt{xyz})^2} = 2x + \sqrt{xyz}\)

Chứng minh tương tự, ta sẽ có:

\(\sqrt{y(4-z)(4-x)}=2y+\sqrt{xyz};\sqrt{z(4-x)(4-y)}=2z+\sqrt{xyz}\)

\(\sqrt{x(4-y)(4-z)}+\sqrt{y(4-z)(4-x)}+\sqrt{z(4-x)(4-y)}-\sqrt{xyz}\)

\(=(2x+\sqrt{xyz})+(2y+\sqrt{xyz})+(2z+\sqrt{xyz})-\sqrt{xyz}\)

\(=2x+2y+2z+2\sqrt{xyz}\)

\(=2(x+y+z+\sqrt{xyz})=2\cdot4=8\) (ĐPCM)

\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)+9\sqrt{x}-4+\left(4x-4\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\)

\(=\dfrac{x+4\sqrt{x}+4x\sqrt{x}+16x-4x-16\sqrt{x}}{x-16}\)

\(=\dfrac{13x+4x\sqrt{x}-12\sqrt{x}}{x-16}\)