Question

1. Phân tích...nhân tử:

a) 2a ( x - y ) - ( y - x )

b) a² ( x - y ) - ( y - x )

c) x ( x - y ) + y ( y - x ) - 3 ( x - y )

d) 6x² - 3 + 7x ( 6x² - 3 ) + 4y ( 3 - 6x² )

2. Tìm x:

a) x² + 2x = 0

b) x ( x - 5 ) = 5 - x

c) ( x + 1 ) ( 6x² + 2x ) + ( x - 1 ) ( 6x² + 2x ) = 0

Answer 1

2 .tìm x

a , x ( x + 2 ) = 0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b, x ( x-5 )= 5 -x

<=> x ( x-5 ) + x - 5 = 0

<=> x (x-5) + ( x-5)= 0

<=> (x-5)(x+1 )=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

c) ( x + 1 ) ( 6x² + 2x ) + ( x - 1 ) ( 6x² + 2x ) = 0

\(\Leftrightarrow\) ( 6x² + 2x ) \([\)(x+1)(x-1)\(]\)=0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x\left(3x+1\right)=0\\x^{2^{ }}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3x+1=0\\x^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\frac{-1}{3}\\x=1\end{matrix}\right.\)

Answer 2

1 ,a) 2a ( x - y ) - ( y - x ) = 2ax - 2ay - y + x

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

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

b) a² ( x - y ) - ( y - x ) = a²x - a²y - y + x

= x ( a²+ 1 ) - y ( a² +1 )

= ( a²+1 ) - (x-y )

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

= x² - 2xy + y² -3x + 3y

= (x-y)² -3 ( x - y )

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