a) 20x - 5y

= 5(4x - y)

b) 5x(x - 1)- 3x(x - 1)

= 2x(x - 1)

c) x(x + y) - 6x - 6y

= x(x + y) - (6x + 6y)

= x(x + y) - 6(x + y)

= (x + y)(x - 6)

d) 6x³ - 9x²

= 3x²(2x - 3)

e) 4x²y - 8xy² + 10x²y²

= 2xy(2x - 4y + 5xy)

g) 20x²y - 12x³

= 4x²(5y - 3x)

h) 8x⁴ + 12x²y⁴ - 16x³y⁴

= 4x²(2x² + 3y⁴ - 4xy⁴)

k) 4xy² + 8xyz

= 4xy(y + 2z)

giúp mình vs ạ c ơn nhiều

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

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

d.

\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

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

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

a, 20x - 5y = 5(4x - y)
b, 4x² - 8xy² + 10x²y = 2x(2x - 4y² + 5xy)
c, 5x (x - 1) - 3x (x - 1) = (5x - 3x) (x - 1) = 2x (x - 1)
d, x (x + y) - 6x - 6y = x (x + y) - (6x + 6y) = x (x + y) - 6 (x + y) = (x - 6) (x + y)
e, x⁴ - y⁴ = (x²)² - (y²)² = (x² - y²) (x² + y²) = [(x + y) (x - y)] (x² + y²)
f, x² - 4y² = x² - (2y)² = (x - 2y) (x + 2y)
g, 27x³ - 64 = (3x)³ - 4³ = (3x - 4) (9x² + 12x + 16)
h, (x +1)² - 16 = (x +1)² - 4² = (x + 1 + 4) (x + 1 - 4) = (x + 5) (x - 3)
i, (3x + 1)² - (x - 2)² = (3x + 1 - x + 2) (3x + 1 + x - 2) = (2x + 3) ( 4x - 1)

`@` `\text {Ans}`

`\downarrow`

^{*Máy tớ cam hơi mờ, cậu thông cảm ._.*}

Cậu viết lại rõ đề câu c, nhé.

Bạn viết đề cẩn thận bằng công thức toán thì sẽ tăng khả năng nhận được sự giúp đỡ hơn. Viết như thế này nhìn rối mắt cực. 

a) 6x³ : (-3x²) = [6 : (-3)] . (x³ : x²)

= -2x

b) (-9x²) : 6x

= (-9 : 6) . (x² : x)

= -3/2 x

c) (-16x⁴) : (-12x³)

= [-16 : (-12)] . (x⁴ : x³)

= 4/3 x

d) (8x³ + 4x² - 6x) : 2x

= 8x³ : 2x + 4x² : 2x - 6x : 2x

= 4x² + 2x - 3

1.a)\(20x-5y=5\left(4x-y\right)\)

b)\(5x\left(x-1\right)-3x\left(x-1\right)=\left(5x-3x\right)\left(x-1\right)=2x\left(x-1\right)\)

c)\(x\left(x+y\right)-6x-6y=x\left(x+y\right)-6\left(x+y\right)=\left(x-6\right)\left(x+y\right)\)

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

e)\(4x^2y-8xy^2+10x^2y^2=2xy\left(2x-8y+10xy\right)\)

g)\(20x^2y-12x^3=4x^2\left(5y-3x\right)\)

h)\(8x^4+12x^2y-16x^3y^4=4x^2\left(2x^2+12y-16xy^4\right)\)

2.a)\(3x\left(x+1\right)-5y\left(x+1\right)=\left(3x-5y\right)\left(x+1\right)\)

b)\(3x\left(x-6\right)-2\left(x-6\right)=\left(3x-2\right)\left(x-6\right)\)

c)\(4y\left(x-1\right)-\left(1-x\right)=4y\left(x-1\right)+\left(x-1\right)=\left(4y+1\right)\left(x-1\right)\)

d)\(\left(x-3\right)^3+3-x=\left(x-3\right)^3-\left(x-3\right)=\left(x-3\right)\left[\left(x-3\right)^2-1\right]=\left(x-3\right)\left(x-2\right)\left(x-4\right)\)

e)\(7x\left(x-y\right)-\left(y-x\right)=7x\left(x-y\right)+\left(x-y\right)=\left(7x+1\right)\left(x-y\right)\)

h)\(3x^3\left(2y-3z\right)-15x\left(2y-3z\right)^2=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]\)

k)Sai đề: \(3x\left(z+2\right)+5\left(-z-2\right)=3x\left(z+2\right)-5\left(z+2\right)=\left(3x-5\right)\left(z+2\right)\)

l)\(18x^2\left(3+x\right)+3\left(x+3\right)=3\left(x+3\right)\left(6x^2+1\right)\)

m)\(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)

n)\(10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)

3.a)\(4x\left(x+1\right)=8\left(x+1\right)\)

\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow4\left(x+1\right)\left(x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

b)\(x\left(x-1\right)-2\left(1-x\right)=0\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

c)\(2x\left(x-2\right)-\left(2-x\right)^2=0\)

\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(2x-x+2\right)\left(x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

d)\(\left(x-3\right)^3+3-x=0\)

\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x-4\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\x-2=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\\x=4\end{matrix}\right.\)

e)\(5x\left(x-2\right)-\left(2-x\right)=0\)

\(\Leftrightarrow5x\left(x-2\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(5x+1\right)\left(x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}5x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\x=2\end{matrix}\right.\)

g)\(5x\left(x-2000\right)-x+2000=0\)

\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\x=2000\end{matrix}\right.\)

h)\(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

k)\(\left(1-x\right)^2-1+x=0\)

\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(1-x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}1-x=0\\-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

m)\(x+6x^2=0\)

\(\Leftrightarrow x\left(1+6x\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\1+6x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{1}{6}\end{matrix}\right.\)

n)\(\left(x+1\right)=\left(x+1\right)^2\)

\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Bạn nhớ ghi vậy x cần tìm là ... nha

a, \(\left(3x-6y\right)x+y\left(x-2y\right)=\left(3x+y\right)\left(x-2y\right)\)

b, \(3\left(x-y\right)-5x\left(y-x\right)=\left(3+5x\right)\left(x-y\right)\)

c, \(\dfrac{2}{5}x\left(y-1\right)+\dfrac{2}{5}y\left(1-y\right)=\left(\dfrac{2}{5}x-\dfrac{2}{5}y\right)\left(y-1\right)=\dfrac{2}{5}\left(x-y\right)\left(y-1\right)\)

d, \(x\left(x-1\right)-y\left(1-x\right)=\left(x+y\right)\left(x-1\right)\)

f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)

g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)

h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)

n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)

p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)

m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)