a,x^2-y^2-2x-2y

= (x-y).(x+y) - 2.(x-y)

= (x-y).(x+y-2)

 b,x^4+2x^3-4x-4

= x^4 - 2x^2 + 2x^3 - 4x + 2x^2 - 4

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

= x^2.(x^2 - 2) + 2x.(x^2 - 2) + 2. (x^2 - 2)

= (x^2 - 2).(x^2 + 2x + 2)

\(a,=2xy\left(2y-x\right)\\ b,=x^2\left(x-4\right)+5\left(x-4\right)=\left(x^2+5\right)\left(x-4\right)\\ c,=\left(x-y\right)\left(x^2-25\right)=\left(x-y\right)\left(x-5\right)\left(x+5\right)\)

a: 6x-2y=2(3x-y)

b: =(x-y)(x-2)(x+2)

Lời giải:
a. Không phân tích được nữa

b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$

$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$

ko phân tích dc

b: =(x-y)(x-2)(x+2)

Lời giải:

a. Vì $x^2\geq 0$ với mọi $x\in\mathbb{R}$ nên $x^2+2\geq 2$

$\Rightarrow A=\frac{32}{x^2+2}\leq \frac{32}{2}=16$

Vậy $A_{\max}=16$ khi $x^2=0\Leftrightarrow x=0$

b.

$(x+1)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow 2(x+1)^2+3\geq 3$

$\Rightarrow B=\frac{5}{2(x+1)^2+3}\leq \frac{5}{3}$

Vậy $B_{\max}=\frac{5}{3}$ khi $x+1=0\Leftrightarrow x=-1$

a: =(xy-2x)-(y^2-2y)

=x(y-2)-y(y-2)

=(x-y)(y-2)

b: =(x^2-2xy+y^2)-(x-y)

=(x-y)^2-(x-y)

=(x-y)(x-y-1)

c: =(x^2-1)-(2xy-2y)

=(x-1)(x+1)-2y(x-1)

=(x-1)(x+1-2y)

d: =(x+3)(x+3-2x+5)

=(x+3)(8-x)

\(a,xy-2x-y^2+2y\)

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

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

\(b,x^2-2xy+y^2-x+y\)

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

\(=\left(x-y\right)\left(x-y-1\right)\)

\(c,x^2-1-2xy+2y\)

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

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

\(d,\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right)\)

\(=\left(x+3\right)\left(x+3-2x+5\right)\)

\(=\left(x+3\right)\left(-x+8\right)\)

#Urushi

a, \(cos3x^2\in\left[-1;1\right]\)

\(\Rightarrow1-cos3x^2\in\left[0;2\right]\)

\(\Rightarrow\sqrt{1-cos3x^2}\in\left[0;\sqrt{2}\right]\)

\(\Rightarrow y=\sqrt{1-cos3x^2}-2\in\left[-2;\sqrt{2}-2\right]\)

\(\Rightarrow y_{min}=-2\Leftrightarrow cos3x^2=1\Leftrightarrow3x^2=k2\pi\Leftrightarrow x=\pm\sqrt{\dfrac{k2\pi}{3}}\)

b, ĐK: \(x\ge1\)

\(cos\sqrt{x-1}\in\left[-1;1\right]\)

\(\Rightarrow y=2008cos\sqrt{x-1}\in\left[-2008;2008\right]\)

\(\Rightarrow y_{min}=-2008\Leftrightarrow cos\sqrt{x-1}=-1\Leftrightarrow\sqrt{x-1}=\pi+k2\pi\Leftrightarrow x=1+\left(\pi+k2\pi\right)^2\)

\(y_{max}=2008\Leftrightarrow cos\sqrt{x-1}=1\Leftrightarrow\sqrt{x-1}=k2\pi\Leftrightarrow x=1+4k^2\pi^2\)

a) \(4\left(2-x\right)^2+xy-2y\)

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

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

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

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

\(=\left(x-2\right)\left(5x-10\right)\)

\(=5\left(x-2\right)^2\)

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

b, \(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-xy+y^2-y^2\right]=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)=x\left(x-y\right)\left(x^2-2xy+y^2-y\right)\)

c, \(=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\)

d, không phân tích được

c, x²y - xy² - 3x + 3y
= xy(x-y) - 3(x-y)
= (x-y)(x-3)

a: =2(x-2)+y(x-2)

=(x-2)(2+y)

b: \(=\left(x+y\right)^2-4=\left(x+y+2\right)\left(x+y-2\right)\)

c: =(x-7)(x+2)

a.

2x - 4 + xy - 2y

= 2(x-2) +y(x-2)

= (x-2)(y+2)

c.

x^2 - 5x - 14

= x^2 + 2x - 7x - 14

= x(x+2) - 7(x+2)

= (x-7)(x+2)

a) Ta có: \(4\left(x-2\right)^2+xy-2y\)

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

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

b) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)

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

\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)

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

\(3x^3+6x+3-3y^2=3\left[\left(x^2+2x+1\right)-y^2\right]=3\left[\left(x+1\right)^2-y^2\right]=3\left(x-y+1\right)\left(x+y+1\right)\)