Bài 2:

\(\left(5x+1\right)^2-\left(2xy-3\right)^2\)

\(=25x^2+10x+1-\left(2xy-3\right)^2\)

\(=25x^2+10x+1\left(4x^2y^2-12xy+9\right)\)

\(=25x^2+10x+1-4x^2y^2+12xy-9\)

\(=25x^2-4x^2y^2+10x+12xy-8\)

Bài 2: 

\(\left(x-1\right)\left(x^2+x+1\right)=x^2\left(x-9\right)+2x+6\)

\(=x^3-1=x^3-9x^2+2x+6\)

\(=x^3-9x^2+2x+6=x^3-1\)

\(=x^3-9x^2+2x+6+1=x^3-1+1\)

\(=x^3-9x^2+2x+7=x^3\)

\(=x^3-9x^2+2x+7-x^3=x^3-x^3\)

\(=-9x^2+2x+7=0\)

\(\Rightarrow x=-\frac{7}{9};x=1\)

 1)    \(25x^4-10x^2y+y^2\)

\(\Leftrightarrow\left(5x^2\right)^2+2\cdot\left(5x^2\right)\cdot y+y^2\)

\(\Leftrightarrow\left(5x^2+y\right)^2\)

 2)   \(x^4+2x^3-4x-4\)

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

 \(\Leftrightarrow\left(x^2-2\right)\left(x^2+2+2x\right)\)

 3)  \(x^4+x^2+1\)

\(\Leftrightarrow x^4+x^2-x+x+1\)

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

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

 4)    \(x^3-5x^2-14x\)\(\Leftrightarrow x^3-7x^2+2x^2-14x\)

\(\Leftrightarrow x^2\left(x-7\right)+2x\left(x-7\right)\)\(\Leftrightarrow x\left(x+2\right)\left(x-7\right)\)

 5)  \(x^2yz+5xyz-14yz\)\(\Leftrightarrow yz\left(x^2+5x-14\right)\)

\(\Leftrightarrow yz\left(x^2+7x-2x-14\right)\)

\(\Leftrightarrow yz\left[x\left(x+7\right)-2\left(x+7\right)\right]\) 

\(\Leftrightarrow yz\left(x+7\right)\left(x-2\right)\)

Cảm ơn bạn Nguyễn Kim Thương :))

tách nhỏ câu hỏi ra

1. -3(-x+3)

= 3x - 6

2. -5x³ (-3x + 5)

= 15x⁴ - 25x³

3. -2x (-2x - 6)

= 4x² + 12x

Giải tiêu biểu câu a nhé.

a/ \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)

\(\Leftrightarrow19x+5=0\)

\(\Leftrightarrow x=-\frac{5}{19}\)

cần câu mấy

Câu mấy cũng được bạn, nếu chỉ có 1 câu thôi thì mình sẽ dựa vào câu đó rồi làm những câu khác, vì mình đang gấp nên giải được hết thì càng tốt, nếu giải được mong bạn ghi chi tiết ra giúp mình^^

b)=x^3+x^2+4x^2+4x+4x+4=x(x+1)+4x(x+1)+4(x+1)=(x+1)(x+4x+4)

Thiên Ân ơi, bạn giải giúp mình câu c được kh

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) \(16x^2y-4xy^2-4x^3+x^2y\)

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

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

\(a)\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}=\frac{-3}{4}\left(x\ne-3;x\ne2\right)\)

\(\Leftrightarrow\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}=\frac{-3}{4}\)

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

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

\(\Leftrightarrow\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)

\(\Leftrightarrow\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)

\(\Leftrightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)

<=> 4x-16=-3x+6

<=> 4x-16+3x-6=0

<=> 7x-22=0

<=> 7x=22

<=> \(x=\frac{22}{7}\)(TMĐK)