1.

\(\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

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

2.

a)  \(27x^4-8x=x\left(27x^3-8\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b)  \(16x^2y-4xy^2-4x^3+x^2y\)

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

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

c) \(x^2-2x-5+2\sqrt{5}\)

\(=\left(x-1\right)^2-6+2\sqrt{5}\)

\(=\left(x-1\right)^2-\left(6-2\sqrt{5}\right)=\left(x-1\right)^2-\left(\sqrt{5}-1\right)^2\)

\(=\left(x-\sqrt{5}\right)\left(x-2+\sqrt{5}\right)\)

Bài 1:

 \(\left(25x^4y^3-15x^3y^5+20x^2y^4\right):\left(5x^2y^3\right)\)

\(=\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

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

Bài 2: 

a) \(27x^4-8x\)

\(=x\left(3x-2\right)\left(3^2x^2+2.3x+2^2\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b) \(16x^2y-4xy^2-4x^3+x^2y\)

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

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

ns thật vs c tôi ms đọc đề bài thôi đã ko hiểu j rồi ns chi đến lm giúp c. Sr nhé

hihi, toán NC mà ms lên đây hỏi

a) 1/2(x³+8)=1/2(x+2)(x²-2x+4)

b) x⁴(x-y)+2x³(x-y)=x³(x+2)(x-y)

c) x²-(y²-6y+9)=x²-(y-3)²=(x-y+3)(x+y-3)

d) xy(x³+y³)=xy(x+y)(x²-xy+y²)

e)3x²(x²-25y²)=3x²(x-5y)(x+5y)

f) 4x⁴+4x²y²+y⁴-4x²y²= (2x²+y²)²-(2xy)²=(2x²-2xy+y²)(2x²+2xy+y²)

a) \(\frac{1}{2}x^3+4=\frac{1}{2}\left(x^3+8\right)=\frac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)

b) \(x^5-x^4y+2x^4-2x^3y=x^3\left(x^2-xy+2x-2y\right)=x^3\left[x\left(x-y\right)+2\left(x-y\right)\right]=x^2\left(x-y\right)\left(x+2\right)\)

c) \(x^2-y^2+6y-9=x^2-\left(y-3\right)^2=\left(x+y-3\right)\left(x-y+3\right)\)

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

e) \(3x^4-75x^2y^2=3x^2\left(x^2-25y^2\right)=3x^2\left(x+5y\right)\left(x-5y\right)\).

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

Mình biết hơi muộn

\(A=x^2+2xy+6x+6y+2y^2+8\Leftrightarrow x^2+2xy+6x+6y+y^2+9-1\)

\(A=0\Rightarrow\left(x+y+3\right)^2+y^2-1=0\)

\(\Rightarrow-1\le x+y+3\le1\) .

\(\Rightarrow2012\le x+y+3+2013\le2014\)

\(\Rightarrow2012\le B\le2014\)

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

Ta có: \(\left(x-2\right)^2\ge0\Rightarrow4-\left(x-2\right)^2\le4\)

\(\Rightarrow A_{max}=4\Leftrightarrow x=2\)

-3;-2;1

\(x^3+4x^2+x-6=0\)

\(\Rightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)

Th1 : \(x-1=0\Rightarrow x=1\)

Th2 : \(x+2=0\Rightarrow x=-2\)

Th3 : \(x+3=0\Rightarrow x=-3\)

bạn có thể trả lời chi tiét hơn ko