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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
M = x9 - x7 + x6 - x5 - x4 + x3 - x2 + 1
= ( x9 - x7 ) + ( x6 - x4 ) - ( x5 - x3 ) - ( x2 - 1 )
= x7( x2 - 1 ) + x4( x2 - 1 ) - x3( x2 - 1 ) - ( x2 - 1 )
= ( x2 - 1 )( x7 + x4 - x3 - 1 )
= ( x - 1 )( x + 1 )[ x4( x3 + 1 ) - ( x3 + 1 ) ]
= ( x - 1 )( x + 1 )( x3 + 1 )( x4 - 1 )
= ( x - 1 )( x + 1 )( x + 1 )( x2 - x + 1 )( x2 - 1 )( x2 + 1 )
= ( x + 1 )2( x - 1 )( x2 - x + 1 )( x - 1 )( x + 1 )( x2 + 1 )
= ( x + 1 )3( x - 1 )2( x2 + 1 )( x2 - x + 1 )
Ta có : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(\left(x+1\right)\left(x+4\right)\right)\left(\left(x+2\right)\left(x+3\right)\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
- Đặt \(x^2+5x+5=a\)
\(=\left(a-1\right)\left(a+1\right)-24=a^2-1-24=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)-96\)
\(=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
\(=\left(x^2+5x+16\right)\left(x+6\right)\left(x-1\right)\)
