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1. \(4x^2-2x-3y-9y^2\)
\(=\left(2x\right)^2-\left(3y\right)^2-\left(2x+3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)-\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x-3y-1\right)\)
2. \(x^2-25=6x-9\)
\(\Rightarrow x^2-6x+9=25\)
\(\Rightarrow\left(x-3\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\)
Bài 2
\(a,x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(b,xy+y^2-x-y\)
\(=y.\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
bài 3
\(a,3x.\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)
vậy x=0,x=2 hay x=-2
\(b,xy+y^2-x-y=0\)
\(y.\left(x+y\right)-\left(x+y\right)=0\)
\(\left(y-1\right).\left(x+y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\x+y=0\end{cases}\Rightarrow\orbr{\begin{cases}y=1\\x=-1\end{cases}}}\)
vậy x=-1, y=1
\(a,x^2+4y^2-4xy\)
\(\Rightarrow\)\(x^2-4xy+\left(2y\right)^2\)
\(\Rightarrow\)\(\left(x-2y\right)^2\)
a, x2 +4y2 -4xy = x2 - 4xy +4y2 = (x - 2y)2
b, x2 y4 +1 - 2xy2 - 9 = x2y4 - 2xy2 +1 -9 =( x2y4 -2xy2 +1)
= (xy2 -1 )2 - 9 =(xy2 -1+3)(xy2 - 1-3)
c, x2- 4x -3 = x2 - 4x +4 - 7
= ( x - 2)2 -7
d, C1 : x2 -8x + 7 = x2 -x -7x +7
= (x2 - x) - (7x +7)
= x(x-1) - 7(x-1)
= ( x - 1)(x - 7)
C2 : x2 - 8x + 7
= x2 - 8x + 16 - 9
= (x2 - 8x +16) -9
= (x - 4 )2 -9
= ( x - 4 +3 )(x - 4 -3 )
=( x - 1 ) (x - 7 )
Good luck !
Bn ko hiểu j cứ hỏi mik nhé !
\(=x^2+x-3x-3.=x\times\left(x+1\right)-3\times\left(x+1\right)=\left(x+1\right).\left(x-3\right)\)
\(x^2-2x-3\)
\(=x^2-3x+x-3\)
\(=x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(x+1\right)\)