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1) x3 - 3x2 = 0
<=> x2( x - 3 ) = 0
<=> \(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
2) 5x( x - 2020 ) - x + 2020 = 0
<=> 5x( x - 2020 ) - ( x - 2020 ) = 0
<=> ( x - 2020 )( 5x - 1 ) = 0
<=> \(\orbr{\begin{cases}x-2020=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{1}{5}\end{cases}}\)
3) ( 3x - 5 )2 = ( x + 1 )2
<=> ( 3x - 5 )2 - ( x + 1 )2 = 0
<=> [ ( 3x - 5 ) - ( x + 1 ) ][ ( 3x - 5 ) + ( x + 1 ) ] = 0
<=> ( 3x - 5 - x - 1 )( 3x - 5 + x + 1 ) = 0
<=> ( 2x - 6 )( 4x - 4 ) = 0
<=> \(\orbr{\begin{cases}2x-6=0\\4x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
4) ( x2 - 2x )2 - 2( x - 1 )2 + 2 = 0
<=> ( x2 - 2x )2 - 2( x2 - 2x + 1 ) + 2 = 0
<=> ( x2 - 2x )2 - 2x2 + 4x - 2 + 2 = 0
<=> ( x2 - 2x )2 - 2( x2 - 2x ) = 0
<=> ( x2 - 2x )( x2 - 2x - 2 ) = 0
<=> \(\orbr{\begin{cases}x^2-2x=0\\x^2-2x-2=0\end{cases}}\)
+) x2 - 2x = 0 <=> x( x - 1 ) = 0 <=> \(\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
+) x2 - 2x - 2 = 0
<=> x2 - 2x + 1 - 3 = 0
<=> ( x2 - 2x + 1 ) = 3
<=> ( x - 1 )2 = ( ±√3 )2
<=> \(\orbr{\begin{cases}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}\)
1/ \(4x^2-12xy+9y^2=\left(2x\right)^2-2.2.3xy+\left(3y\right)^2\)
\(=\left(2x-3y\right)^2\)
2/ \(x^3-y^6=x^3-\left(y^2\right)^3\)
\(=\left(x-y^2\right)\left(x^2+xy^2+y^4\right)\)
Làm tạm 2 phần đợi mik xíu
4x2 - 12xy + 9y2 = ( 2x )2 - 2.2x.3y + ( 3y )2 = ( 2x - 3y )2
x3 - y6 = x3 - ( y2 )3 = ( x - y2 )( x2 + xy2 + y4 )
x6 - 6x4 + 12x2 - 8 = ( x2 )3 - 3.(x2)2.2 + 3.x2.22 - 23 = ( x2 - 2 )3
( x2 + 4y2 - 5 )2 - 16( x2y2 + 2xy + 1 ) = ( x2 + 4y2 - 5 )2 - 42( xy + 1 )2
= ( x2 + 4y2 - 5 )2 - ( 4xy + 4 )2
= [ ( x2 + 4y2 - 5 ) - ( 4xy + 4 ) ][ ( x2 + 4y2 - 5 ) + ( 4xy + 4 ) ]
= ( x2 + 4y2 - 5 - 4xy - 4 )( x2 + 4y2 - 5 + 4xy + 4 )
= [ ( x2 - 4xy + 4y2 ) - 9 ][ ( x2 + 4xy + 4y2 ) - 1 ]
= [ ( x - 2y )2 - 32 ][ ( x + 2y )2 - 12 ]
= ( x - 2y - 3 )( x - 2y + 3 )( x + 2y - 1 )( x + 2y + 1 )
( a + b )3 - ( a3 + b3 ) = a3 + 3a2b + 3ab2 + b3 - a3 - b3
= 3a2b + 3ab2
= 3ab( a + b )
+) \(ax-a+bx-b+x-1=a\left(x-1\right)+b\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(a+b+1\right)\)
+) Xem lại đề
ax - a + bx - b + x - 1
= a( x - 1 ) + b( x - 1 ) + 1( x - 1 )
= ( x - 1 )( a + b + 1 )
x3 - 2x2 - 2x + 4 ( sửa -4 thành +4 )
= x2( x - 2 ) - 2( x - 2 )
= ( x - 2 )( x2 - 2 )
Bonus = ( x - 2 )[ x2 - ( √2 )2 ]
= ( x - 2 )( x - √2 )( x + √2 )
Phân tích đa thức thành nhân tử à?
1) \(\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
2) \(x^3+1-x^2-x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left[x^2-x+1-x\right]\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
( x + y )3 - x3 - y3
= ( x + y )3 - ( x3 + y3 )
= ( x + y )3 - ( x + y )( x2 - xy + y2 )
= ( x + y )[ ( x + y )2 - ( x2 - xy + y2 ) ]
= ( x + y )( x2 + 2xy + y2 - x2 + xy - y2 )
= 3xy( x + y )
x3 + 1 - x2 - x
= ( x3 + 1 ) - ( x2 + x )
= ( x + 1 )( x2 - x + 1 ) - x( x + 1 )
= ( x + 1 )( x2 - x + 1 - x )
= ( x + 1 )( x2 - 2x + 1 )
= ( x + 1 )( x - 1 )2