a

\(xy+3x-7y-21\\ =\left(xy+3x\right)-\left(7y+21\right)\\ =x\left(y+3\right)-7\left(y+3\right)\\ =\left(y+3\right)\left(x-7\right)\)

b

\(2xy-15-6x+5y\\ =\left(2xy-6x\right)-\left(15-5y\right)\\ =2x\left(y-3\right)-5\left(3-y\right)\\ =2x\left(y-3\right)+5\left(y-3\right)\\ =\left(y-3\right)\left(2x+5\right)\)

c Đề phải là \(\left(2x^2y+2xy^2-x-y\right)\) mới phân tích được: )

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

d

\(7x^3y-3xyz-21x^2+9z\\ =\left(7x^3y-21x^2\right)-\left(3xyz-9z\right)\\ =7x^2\left(xy-3\right)-3z\left(xy-3\right)\\ =\left(xy-3\right)\left(7x^2-3z\right)\)

e

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

f

\(9x^2-25y^2-6x+10y\\ =\left(3x\right)^2-\left(5y\right)^2-\left(6x-10y\right)\\ =\left(3x-5y\right)\left(3x+5y\right)-2\left(3x-5y\right)\\ =\left(3x-5y\right)\left(3x+5y-2\right)\)

a: =x(y+3)-7(y+3)

=(y+3)(x-7)

b: \(=2xy-6x+5y-15\)

=2x(y-3)+5(y-3)

=(y-3)(2x+5)

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

d: \(=xy\left(7x^2-3z\right)-3\left(7x^2-3z\right)\)

=(7x^2-3z)(xy-3)

e: =4x^2-y^2-2x-y

=(2x-y)(2x+y)-(2x+y)

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

f: =(3x-5y)(3x+5y)-2(3x-5y)

=(3x-5y)(3x+5y-2)

a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4 
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))

Bài 1:

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

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

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

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

a) \(2x-2y-x^2+2xy-y^2\)

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

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

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

b) \(9x^2+6xy+y^2-25\)

\(=\left(3x\right)^2+6xy+y^2-25\)

\(=\left(3x+y\right)^2-5^2\)

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

a,2x²+9x-35

=2x²+14x-5x-35

=2x(x+7)-5(x+7)

=(x+7)(2x-5)

b,12+x-6x²

=12-9x+8x-6x²

=3(4-3x)+2x(4-3x)

=(4-3x)(3+2x)