a) \(\frac{x+2}{-5}\ge0\Leftrightarrow x+2\le0\Leftrightarrow x\le-2\)

b) Điều kiện : \(x\ne3\)

\(\frac{x-1}{x-3}>1\Leftrightarrow\frac{x-1-x+3}{x-3}>0\Leftrightarrow\frac{2}{x-3}>0\Leftrightarrow x-3>0\Leftrightarrow x>3\)

Vậy BĐT luôn đúng với mọi \(x>3\)

a)\(\frac{x+2}{-5}\ge0\Leftrightarrow x+2\ge0\Leftrightarrow x\ge-2\)

b)\(\frac{x-1}{x-3}>1\Leftrightarrow\frac{x-1}{x-3}-1>0\Leftrightarrow\frac{2}{x-3}>0\Leftrightarrow x=\frac{2}{0}+3\)=> vô nghiệm 

ĐKXD: X khác 3

a) \(6x^2+6\)

\(=6\left(x^2+1\right)\)

b) \(2x^2-18\)

\(=2\left(x^2-9\right)\)

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

c) \(3x^2-3xy+4x-4y\)

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

\(=3x\left(x-y\right)+4\left(x-y\right)\)

\(=\left(3x-4\right)\left(x-y\right)\)

a) \(\left(x^3-9x^2+27x-27\right)\)\(:\)\(\left(x-3\right)\)

\(=\left(x-3\right)^3\)\(:\)\(\left(x-3\right)\)

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

c) \(\frac{x^2-4}{2x}:\frac{3x-6}{6}\)

\(=\frac{\left(x-2\right)\left(x+2\right)}{2x}.\frac{6}{3\left(x-2\right)}\)

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

\(a,x^3-3.x^2-4x+12\)

\(=\left(x^3-3x^2\right)-\left(4x-12\right)\)

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

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

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

\(a,x^3-3x^2-4x+12\)

\(=\left(x^3-3x^2\right)-\left(4x-12\right)\)

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

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

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

1)

$x^3+9x^2+23x+15=(x^3+x^2)+(8x^2+8x)+(15x+15)$

$=x^2(x+1)+8x(x+1)+15(x+1)$

$=(x+1)(x^2+8x+15)$

$=(x+1)[(x^2+3x)+(5x+15)]$

$=(x+1)[x(x+3)+5(x+3)]=(x+1)(x+3)(x+5)$

5)

$x^4+5x^2+9=(x^4+6x^2+9)-x^2$

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

3)

$(3x-2)^2(6x-5)(6x-3)-5$

$=(9x^2-12x+4)(36x^2-48x+15)-5$

$=(9x^2-12x+4)[4(9x^2-12x)+15]-5$

$=(a+4)(4a+15)-5$ (đặt $9x^2-12x=a$)

$=4a^2+31a+55$

$=4a^2+20a+11a+55$

$=4a(a+5)+11(a+5)=(4a+11)(a+5)=(36x^2-48x+11)(9x^2-12x+5)$

$=