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

a)(x³- x) - (2x² - 2)

= x (x² - 1) - 2 (x² - 1)

= (x - 2) (x²-1)

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

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

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

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

a, = (xy-y^2) + (2x-2y) = y(x-y) + 2.(x-y) = (x-y).(y+2) 

b, = (x+y)^2 - 9 = (x+y-3).(x+y+3)

\(xy-y^2-2x-2y\)   

=(xy-y).(2x-2y)

=y(x-y).2(x-y)

=(x-y).(y-2)

camr ơn bạn 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

(*)\(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(x-3\right)\left(3x-2\right)\)

(*)\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-5\right)\left(x-1\right)\)

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

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

(*)\(6x^2+7xy+2y^2=6x^2+4xy+3xy+2y^2=2x\left(3x+2y\right)+y\left(3x+2y\right)=\left(2x+y\right)\left(3x+2y\right)\)

a, \(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(3x-2\right)\left(x-3\right)\)

b, \(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)

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

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

e, \(6x^2+7xy+2y^2=6x^2+3xy+4xy+2y^2=3x\left(2x+y\right)+2y\left(2x+y\right)=\left(2x+y\right)\left(3x+2y\right)\)

\(a,x^2+6x+9\)

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

\(b,10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

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

\(c,8x^3-\frac{1}{8}\)

\(=8x^3-\left(\frac{1}{2}\right)^3\)

\(=\left(8x-\frac{1}{2}\right)\left(64x^2+4x+\frac{1}{4}\right)\)

\(d,8x^3+12x^2+6xy^2+y^3\)

\(=2\left(4x^3+6x^2+3xy^2+\frac{1}{2}y^3\right)\)

hok tốt!

ai tra loi dung minh cho

Điệp viên 007 sai c

c, \(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

Viết đề rõ chút chứ nhìn ko ra

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