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\(\Rightarrow x^2-5x-36=0\Rightarrow x\left(x-9\right)+4\left(x-9\right)=0\Rightarrow\left(x-9\right)\left(x+4\right)=0\Rightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
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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
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\(\Leftrightarrow\left(x-7\right)\left(x^2+x+6\right)-x\left(x-7\right)=0\\ \Leftrightarrow\left(x-7\right)\left(x^2+6\right)=0\\ \Leftrightarrow x=7\left(x^2+6>0\right)\)
x^2-x=20
=>x^2-x-20=0
=>x^2+4x-5x-20=0
=>x(x+4)-5(x+4)=0
=>(x-5)(x+4)=0
còn lại tự làm
cảm ơn bạn