a: =x(x-5)

a) \(=x\left(x-5\right)\)

b) \(=\left(x+3y-3y\right)\left(x+3y+3y\right)=x\left(x+6y\right)\)

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

b,=(x+3y-3y(x+3y+3y)

= x(x+6y)

a) x²-xz-9y²+3yz

=(x²-9y²)-(xz-3yz)

=(x-3y)(x+3y)-z(x-3y)

=(x-3y)(x+3y-z)

b)x³-x²-5x+125

=x³-6x²+25x+5x²-30x+125

=x(x²-6x+25)+5(x²-6x+25)

=(x+5)(x²-6x+25)

c.x³+2x²-6x-27

=x³+5x²+9x-3x²-15x-27

=x(x²+5x+9)-3(x²+5x+9)

=(x-3)(x²+5x+9)

d. 12x³+4x²-27x-9

=12x³+4x²-27x-9

=4x²(3x+1)-9(3x+1)

=(4x²-9)(3x+1)

=(2x-3)(2x+3)(3x+1)

e.x⁴-25x²+20x-4

=x⁴+5x³-2x²-5x²-25x+10+2x²+10x-4

=x²(x²+5x-2)-5(x²+5x-2)+2(x²+5x-2)

=(x²-5x+2)(x²+5x-2)

f.x²(x²-6)-x²+9

=x⁴+x³-3x²-x³-x²+3x-3x²-3x+9

=x²(x²+x-3)-x(x²+x-3)-3(x²+x-3)

=(x²-x-3)(x²+x-3)

mà bạn ơi sao câu b bạn tách ra nv ?

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

                         = -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - 1/8 = (2x)³ – (1/2)³ = (2x - 1/2)[(2x)² + 2x . 12 + (1/2)²]

                    ^{= (2x - 1/2)(4x2 + x + 1/4)} 

d)1/25x² – 64y² = (1/5x)2(1/5x)2- (8y)² = (1/5x + 8y)(1/5x - 8y)

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

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

b: Ta có: \(xy-3x-2y+6\)

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

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

c: Ta có: \(x^2-6xy+9y^2-4z^2\)

\(=\left(x-3y\right)^2-4z^2\)

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

a. \(x^2\) - 9y²

= (\(x\))² - (3y)²

= (\(x\) - 3y)(\(x\) + 3y)

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

B. 7\(x\).(\(x\) - y) + 3\(x\) - 3y

= 7\(x\).(\(x\) -y) + 3.(\(x\) - y)

= (\(x\) - y)(7\(x\) + 3)

Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)

Lời giải:
a. $5x^2-10xy=5x(x-2y)$

b. $3x(x-y)-6(x-y)=(x-y)(3x-6)=3(x-y)(x-2)$
c. $2x(x-y)-4y(y-x)=2x(x-y)+4y(x-y)=(x-y)(2x+4y)=2(x-y)(x+2y)$

d. $9x^2-9y^2=9(x^2-y^2)=9(x-y)(x+y)$

e. $x^2-xy-x+y=(x^2-xy)-(x-y)=x(x-y)-(x-y)=(x-y)(x-1)$

f. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$

b: Ta có: \(xy-3x-2y+6\)

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

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

\(a,15x-5xy\\ =5x\left(3-y\right)\\ b,\left(x^2+1\right)^2-4x^2\\ =\left(x^2-x+1\right)\left(x^2+x+1\right)\\ c,x^2-10x-9y^2+25\\ =\left(x-5\right)^2-9y^2\\ =\left(x-9y-5\right)\left(x+9y-5\right)\)

hé lu ông zà

a) 5x(3 - y)

b) (x² - x + 1)(x² + x + 1)

c) (x - 9y - 5)(x + 9y - 5)

b) \(x^2y-x^3-10y+10x\)

\(=x^2\left(y-x\right)-10\left(y-x\right)\)

\(=\left(y-x\right)\left(x^2-10\right)\)

c) \(x^2\left(x-2\right)+49\left(2-x\right)\)

\(=\left(x-2\right)\left(x^2-49\right)\)

\(=\left(x-2\right)\left(x-7\right)\left(x+7\right)\)

a) \(A=x^2-6x+9-9y^2\)

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

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

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

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

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

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

a, \(A=\left(x-3\right)^2-9y^2=\left(x-3-3y\right)\left(x-3+3y\right)\)

b, \(B=\left(x-1\right)^3+2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left[\left(x-1\right)^2+2\left(x+1\right)\right]\)

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