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

B) \(\left(5x-6y\right)\left(x-2\right)=5x^2-10x-6xy+12y\)

C) \(\dfrac{10x^4y^2-5x^3y^2+15xy^4}{5xy}=\dfrac{5xy\left(2x^3y-x^2y+3y^3\right)}{5xy}\)

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

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

\(E)4x^2-49\)

F) \(x\left(x+2\right)-2x-4=0\)

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

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

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

\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

A, 2x.(x+1) = 2x\(^2\)+2x B,(5x-6y)(x-2) = 5x\(^2\)-10x-6xy+12y C,(10x\(^4\)y\(^2\)-5x\(^3\)y\(^2\)+15xy\(^4\)) : 5xy = 2x\(^3\)y - x\(^2\)y + 3y\(^3\)

a) \(2x\left(x-7\right)-5y\left(x-7\right)=\left(x-7\right)\left(2x-5y\right)\)

b) \(5x^3y+10x^2y+5xy=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

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

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

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

a: \(=\left(x-7\right)\left(2x-5y\right)\)

b: \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

a) \(=\left(2x-5y\right)\left(x-7\right)\)

b) \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

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

Bài 1: 

a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)

\(=\dfrac{15x^2y^2z}{3xyz}\)

\(=5xy\)

b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)

\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)

\(=15x^4-12x^3+9x^2\)

c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)

\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)

\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)

\(=2x+5+\dfrac{20}{x-4}\)

d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)

\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)

\(=-15x^3y^2+25x^2y^2-5xy^3\)

1, a, \(2x\left(x+1\right)=2x^2+2x\)

b, \(\left(5x-6\right)\left(x-2\right)=5x^2-16x+12\)

c, \(\left(10x^4y^2-5x^3y^2+15xy^2\right):5xy=2x^3y-x^2y+3y\)

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

b, \(4x^2-49=\left(2x\right)^2-7^2=\left(2x+7\right)\left(2x-7\right)\)

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

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

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

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

a) (10x³y - 5x²y² - 25x⁴y³) : (- 5xy)

= - 2x² + xy + 5x³y²

b) (27x³ - y³) : (3x - y)

= (3x - y)(9x² - 3xy + y²) : (3x - y)

= 9x² - 3xy + y²

a) ( 10x³y - 5x²y² - 25 x⁴y³) : ( -5xy)

Ta có : -5xy( -2x² + xy + 5x³y²) : ( - 5xy)

Vậy , ta được thương là : -2x² + xy + 5x³y²

b) ( 27x³ - y³) : ( 3x - y)

Ta có : ( 3x - y)( 9x² + 3xy + y²) : ( 3x - y)

Vậy , ta được thương là : 9x² + 3xy + y²

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

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

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

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

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

Bài 4:

3: 10x=6y=5z

\(\Leftrightarrow\dfrac{10x}{30}=\dfrac{6y}{30}=\dfrac{5z}{30}\)

hay x/3=y/5=z/6

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{24}{2}=12\)

Do đó: x=36; y=60; z=72

4: Ta có: 9x=3y=2z

nên \(\dfrac{9x}{18}=\dfrac{3y}{18}=\dfrac{2z}{18}\)

hay x/2=y/6=z/9

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{6}=\dfrac{z}{9}=\dfrac{x-y+z}{2-6+9}=\dfrac{50}{5}=10\)

Do đó: x=20; y=60; z=90