Thêm điều kiện \(x>5\).

\(\frac{x^2-2x-6}{x-5}=\frac{x^2-5x+3x-15+9}{x-5}=x+3+\frac{9}{x-5}\)

\(=x-5+\frac{9}{x-5}+8\ge2\sqrt{\left(x-5\right).\frac{9}{x-5}}+8\)

\(=2.3+8=14\)

Dấu \(=\)khi \(x-5=\frac{9}{x-5}\Rightarrow x=8\).

x(x+1)² + x(x-5) - 5(x+1)²

= (x+1)² (x -5) + x(x-5)

= [(x+1)² + x](x-5) 

=(x² + 2x + x +1)(x-5)

= (x² + 3x+1)(x-5)

2.  x²(y-z) + y²(z-x) + z²(x-y)

= x²[(y-x) +(x-z)] + y²(z-x) + z² (x-y)

= x²(y-x) + x²(x-z) + y²(z-x) + z²(x-y)

= x²(y-x) + x²(x-z) - y²(x-z) - z²(y-x)

= (y-x)(x²-z²) + (x-z)(x² -y²)

= (y-x)(x-z)(x+z) + (x-z)(x-y)(x+y)

= (x-y)(x-z)(x+y - x -z)

= (x-y)(x-z)(y-z)

a, \(x^3-2x=0\Leftrightarrow x\left(x^2-2\right)=0\Leftrightarrow x=;x=\pm\sqrt{2}\)

b, \(x^2\left(x-3\right)+12-4x=0\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\Leftrightarrow x=\pm2;x=3\)

c, \(\left(x-2\right)^2=x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-x^2-2x-4\right)=0\Leftrightarrow\left(x-2\right)\left(-x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+6>0\right)=0\Leftrightarrow x=2\)

d, \(x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow x=2;x=3\)

e, \(x^3-4x^2+2x-1=0\Leftrightarrow x=3,5...\)

\(4ab-8a^2b=4ab\left(1-2a\right)\)

\(x^2-4+3xy-6y\)

\(=\left(x^2-4\right)+\left(3xy-6y\right)\)

\(=\left(x-2\right)\left(x+2\right)+3y\left(x-2\right)\)

\(=\left(x-2\right)\left(x+2+3y\right)\)

\(x^2-8x+12\)

\(=x^2-6x-2x+12\)

\(=x\left(x-6\right)-2\left(x-6\right)\)

\(=\left(x-2\right)\left(x-6\right)\)

\(x^4+64\)

\(=x^4+16x^2+64-16x^{2^2}\)

\(=\left(x^2+8\right)^2-\left(4x\right)^2\)

\(=\left(x^2-4x+8\right)\left(x^2^2+4x+8\right)\)

\(x^3-3x^2+4x-2\)

\(=x^3-x^2-2x^2+2x+2x-2\)

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

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

\(x^3-3x+4\)

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

\(=\left(x^3-x^2\right)+\left(x^2+x\right)-\left(4x-4\right)\)

\(=x^2\left(x-1\right)+x\left(x+1\right)-4\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x-4\right)\)

\(x^5+x^4+1\)

\(=x^5-x^3-x^2-x^4+x^2+x+x^3-x-1\)

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

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

ko vì tích của 3 số tự nhiên liên tiếp luôn là 1 số chẵn mà 3^50 là số lẻ nên mâu thuẫn.Do đó 3^50 ko là tích của 3 số tự nhiên liên tiếp