\(3x^2y-12x^3y^2=3x^2y\left(1-4xy\right)\)

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

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

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

2) \(x^2+6x+9+x^2y+3xy\)

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

\(=\left(x+3\right)\left(x+xy+3\right)\)

\(\left(2x+3y\right)^2+\left(3x-2y\right)^2-2.\left(2x+3y\right).\left(3x-2y\right)\)

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

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

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

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

Help meee plzzz

em ko lam dc anh ehh

em chua gap bai nao nhu the nay. noi dung hon la chua den lop lam haha

SORRY BUT I CAN'T HELP

Tìm x:

\(8x^2-\left(2x+5\right)\left(4x-2\right)-9=0\)

\(\Leftrightarrow8x^2-\left(8x^2-4x+20x-10\right)-9=0\)

​\(\Leftrightarrow8x^2-8x^2+4x-20x+10-9=0\)​

\(\Leftrightarrow-16x+1=0\)

\(\Leftrightarrow-16x=-1\)

\(\Leftrightarrow x=\dfrac{-1}{-16}=\dfrac{1}{16}\)

Vậy \(x=\dfrac{1}{16}\)

Bài 1:

\(a,8x^2-\left(2x+5\right)\left(4x-2\right)-9=0\)

\(\Rightarrow8x^2-\left(8x^2+16x-10\right)-9=0\)

\(\Rightarrow8x^2-8x^2-16x+10-9=0\)

\(\Rightarrow-16x+1=0\)

\(\Rightarrow x=\dfrac{1}{16}\)

Tìm x:

a) 8x² - (2x + 5)(4x - 2) - 9 =0

8x² - (8x² - 4x + 20x - 10) - 9= 0

8x² - (8x² + 16x - 10) - 9 =0

8x² - 8x² - 16x + 10 - 9 = 0

- 16x + 1 =0

- 16x = - 1

x = \(\frac{1}{16}\)

Vậy \(x = \frac{1}{16}\)

PTĐTTNT:

3/ x² - 3x + 2

= x² - x - 2x + 2

= x(x - 1) - 2(x - 1)

= (x - 1)(x - 2)

pn coi lại câu 1 và 2 nhé