\(x+\sqrt{x}-12=\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

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

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

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

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

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

b,   x²-2xy+y²-z²

=(x-y)²-z²

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

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

A,          m²-7m+12=m²-3m+12-4m=m(m-3)-4(m-3)=(m-4)(m-3)

B,           2x⁴-x³+27-54x=x³(2-x)-27(2-x)=(x³-27)(2-x)=(x-3)(x²+3x+9)(2-x)

a) m^2 -7m +12 = m^2 -3m -4m +12

=m(m -3)-4 (m- 3)

=(m-4)(m-3)

b) 2x^4-x^3 -54x+ 27

=(2m^4-x^3)- (54x - 27)

=x^3(2x-1)-27(2x-1)

=(x^3-27)(2x-1)

a,(m-4)(m-3)

b,(x-3)(2x-1)(x^2-3x+9)

b ( x^2 + 3x + 2)( x^2 + 7x + 12) - 24 

= [ x^2 +x + 2x + 2) ( x^2 +3x + 4x + 12) - 24

= [x(x+1) + 2 (x + 1) [x(x+3) + 4(x+3) ] - 24

= ( x + 1)(x+2) (x+3)(x+4) - 24

= ( x + 1).(x+4) (x+2)(x+3) - 24

=(x^2 + 5x + 4)(x^2+5x+6) - 24 

Đặt x^2 + 5x +4 =y ta có:

= y(y+2) - 24

= y^2 + 2y - 24 

= y^2 + 2y + 1 - 25 

= ( y + 1)^2 - (5)^2 

= ( y + 1 - 5 )( y + 1 + 5)

= ( y- 4)(y +6) 

Thay y trở lại là đc 

đúng nha

Bài 1:

\(x^2-6x+9-y^2\)

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

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

Bài 2:

\(x^2-x-12=0\)

\(\Leftrightarrow x^2-4x+3x-12=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)

1. x²+6x-9-y²

=-(x²-6x+y²)-3²

=-(x-y)²-3²

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

2. x² - x - 12 = 0

=> x² - 4x + 3x - 12 = 0

=> x( x - 4 ) + 3( x - 4 ) = 0

=> ( x + 3 )(x - 4 ) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0 \end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)

Vậy x=-3; x=4

Câu a :

\(\dfrac{3x^2-12x+12}{x^4-8x}\)

\(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)

\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)

Câu b :

\(\dfrac{7x^2+14x+7}{3x^2+3x}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

\(x^3-64=x^3-4^3\)

\(\Rightarrow\left(x-4\right)\left(x^2+4x+4^2\right)\)

x³-64=x³-4³=(x-4)(x²+4x+16)

Ta có:\(x^3-64\)

\(=x^3-4^3\)

Áp dụng hằng đẳng thức:\(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)

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