a. x² - 2x

⇔ x(x - 2)

b. 3x - 6y

⇔ 3(x - 2y)

c. 5(x + 3y) - 15x(x + 3y)

⇔ (5 - 15x)(x + 3y)

d. 3(x - y) - 5x(y - x)

⇔ 3(x - y) + 5x(x - y)

⇔ (3 + 5x)(x - y)

2B: 

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

b: \(x\left(x+y\right)^2+xy\left(x+y\right)=x\left(x+y\right)\left(x+2y\right)\)

c: \(xy\left(x-y\right)-2x+2y=\left(x-y\right)\left(xy-2\right)\)

d: \(y\left(x-y\right)^2+x\left(y-x\right)^2=\left(x-y\right)^2\cdot\left(x+y\right)\)

\(Q=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\)

\(\Leftrightarrow Q=\frac{3x+3y+2z}{\sqrt{6\left(x^2+xy+yz+zx\right)}+\sqrt{6\left(y^2+xy+yz+zx\right)}+\sqrt{z^2+xy+yz+zx}}\)

\(\Leftrightarrow Q=\frac{3x+3y+2z}{\sqrt{3\left(x+y\right).2\left(x+z\right)}+\sqrt{3\left(y+x\right).2\left(y+z\right)}+\sqrt{\left(z+x\right).\left(z+y\right)}}\)

\(\Rightarrow Q\ge\frac{3x+3y+2z}{\frac{3\left(x+y\right)+2\left(x+z\right)}{2}+\frac{3\left(y+x\right)+2\left(y+z\right)}{2}+\frac{\left(z+x\right)+\left(z+y\right)}{2}}\)

\(\Rightarrow Q\ge\frac{3x+3y+2z}{\frac{9x+9y+6z}{2}}=\frac{2}{3}\)

Dấu "=" xảy ra khi \(x=y=1\)và  \(z=2\)

K = a² - 2ab + 5b² - 4b + 9

= (a² - 2ab + b²) + (4b² - 4b + 1) + 8

= (a - b)² + (2b - 1)² + 8

Do (a - b)² ≥ 0 với mọi a, b ∈ R

(2b - 1)² ≥ 0 với mọi b R

⇒ (a - b)² + (2b - 1)² ≥ 0 với mọi a, b ∈ R

⇒ (a - b)² + (2b - 1)² + 8 ≥ 8 với mọi a, b ∈ R

Vậy GTNN của K là 8 khi a = b = 1/2

\(A=\dfrac{x^3-2x^2-15x}{x-5}=\dfrac{x\left(x^2-2x-15\right)}{x-5}=\dfrac{x\left(x+3\right)\left(x-5\right)}{x-5}=x\left(x+3\right)\)

\(A=x^2+3x=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)

\(A_{min}=-\dfrac{9}{4}\)

Ta có: \(E=4x^2+4x-5\)

\(=4x^2+4x+1-6\)

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

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

\(A=4x^2+4x-5=4x^2+4x+1-6=\left(2x+1\right)^2-6\)

Do \(\left(2x+1\right)^2\ge0\) \(\Rightarrow\left(2x+1\right)^2-6\ge-6\)

\(\Rightarrow Max\) A=-6\(\Leftrightarrow x=\dfrac{-1}{2}\)

3n+2 \(⋮\)n-1

=> 3n+1 \(⋮\)n-1

=> (3n +1) - 3(n-1)

=> (3n+1) - ( 3n-3)

=> 3n+1 -3n+3

=> ( 3n-3n) + (1+3)

=> 4 \(⋮\)n-1

=> n-1 \(\in\)Ư(4)= { 1;2 ;4; -1; -2; -4}

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