Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
BT1
a, 3x2 - 9x = 3x( x - 3 )
b, x2 - 4x + 4 = ( x - 2 )2
c, x2 + 6x + 9 - y2 = ( x2 + 6x + 9 ) - y2
= ( x + 3 )2 -y2
= ( x + 3 - y ) ( x + 3 + y )
BT2
a, 1012 - 1 = ( 101 - 1 ) ( 101 + 1 ) = 100 . 102 = 10200
b, 672 + 66 . 67 + 332 = 672 + 2 . 67 . 33 + 332
= ( 67 + 33 )2 = 1002 = 10000
BT3
x ( x - 3 ) + 2 ( x - 3 ) = 0
<=> ( x + 2 )( x - 3 ) = 0
<=> x + 2 = 0 <=> x = -2
x - 3 = 0 x = 3
vậy ................
BT1
a) 3x2 - 9x = 3x( x - 3 )
b) x2 - 4x + 4 = ( x - 2 )2
c) x2 + 6x + 9 - y2
= ( x2 + 6x + 9 ) - y2
= ( x + 3 )2 - y2
= ( x - y + 3 )( x + y + 3 )
BT2.
a) 1012 - 1 = 1012 - 12 = ( 101 - 1 )( 101 + 1 ) = 100.102 = 10 200
b) 672 + 66.67 + 332 = 672 + 2.33.67 + 332 = ( 67 + 33 )2 = 1002 = 10 000
BT3.
x( x - 3 ) + 2( x - 3 ) = 0
<=> ( x - 3 )( x + 2 ) = 0
<=> x - 3 = 0 hoặc x + 2 = 0
<=> x = 3 hoặc x = -2
