a, \(\frac{5}{2}x^2y^2+15x^2y-30xy^2=5xy\left(\frac{1}{2}xy+3x-6y\right)\)

b, \(16x^2+24x-8xy-6y+y^2\)

\(=\left(16x^2-8xy+y^2\right)+\left(24x-6y\right)=\left(4x-y\right)^2+6\left(4x-y\right)\)

\(=\left(4x-y\right)\left[\left(4x-y\right)+6\right]=\left(4x-y\right)\left(4x-y+6\right)\)

c, \(2x^2-5x-7=2x^2-7x+2x-7\)

\(=2x\left(x+1\right)-7\left(x+1\right)=\left(2x-7\right)\left(x+1\right)\)

Bai lam 

\(4\left(x^2-2\right)=0\Leftrightarrow x^2-2=0\)

\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\)

\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)+b\left(c^3-b^3+b^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)+b\left(c^3-b^3\right)+b\left(b^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)

\(=\left(b^3-c^3\right)\left(a-b\right)-\left(a^3-b^3\right)\left(b-c\right)\)

\(=\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-b\right)-\left(a-b\right)\left(a^2+ab+b^2\right)\left(b-c\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(b^2+bc+c^2-a^2-ab-b^2\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(bc+c^2-a^2-ab\right)\)

\(=\left(b-c\right)\left(a-b\right)\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)

\(=\left(b-c\right)\left(a-b\right)\left(c-a\right)\left(a+b+c\right)\)

\(\hept{\begin{cases}f\left(x\right)=x^4+6x^2+25⋮P\left(x\right)\\g\left(x\right)=3x^4+4x^2+28x+5⋮P\left(x\right)\end{cases}}\)

\(\Rightarrow3f\left(x\right)-g\left(x\right)=3\left(x^4+6x^2+25\right)-\left(3x^4+4x^2+28x+5\right)=14\left(x^2-2x+5\right)⋮P\left(x\right)\)

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

Thử lại: \(f\left(x\right)=\left(x^2+2x+5\right)\left(x^2-2x+5\right)⋮P\left(x\right)\)

              \(g\left(x\right)=\left(3x^2+6x+1\right)\left(x^2-2x+5\right)⋮P\left(x\right)\)

do đó \(P\left(x\right)=x^2-2x+5\)thỏa mãn. 

\(P\left(-2\right)=\left(-2\right)^2-2.\left(-2\right)+5=13\)

\(a\left(a-b\right)+b\left(b-c\right)+c\left(c-a\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a=b=c\).

\(A=a^3+b^3+c^3-3abc+3ab-3c+5\)

\(A=a^3+a^3+a^3-3a^3+3a^2-3a+5\)

\(A=3a^2-3a+5\)

\(A=3\left(a-\frac{1}{2}\right)^2+\frac{17}{4}\ge\frac{17}{4}\).

Dấu \(=\)khi \(a=\frac{1}{2}\).

Vậy GTNN của \(A\)là \(\frac{17}{4}\).