mọi người sử dụng định lý bezout để làm dạng 4 giúp mình ạ, mình đang cần rất gấp ạ
Dạng 3:
\(1,=\left(x+y\right)^2:\left(x+y\right)=x+y\\ 2,=\left(5x+1\right)\left(25x^2-5x+1\right):\left(5x+1\right)=25x^2-5x+1\\ 3,=\left(y-x\right)^2:\left(y-x\right)=y-x\\ 4,=\left(2x-3y\right)\left(2x+3y\right):\left(2x-3y\right)=2x+3y\\ 5,=\left(3x-1\right)\left(9x^2+3x+1\right):\left(3x-1\right)=9x^2+3x+1\\ 6,=\left(2x+1\right)\left(4x^2-2x+1\right):\left(4x^2-2x+1\right)=2x+1\\ 7,=\left(x-3\right)\left(x+y\right):\left(x+y\right)=x-3\)
Dạng 4:
\(3,\Leftrightarrow2x^2+ax+a=\left(x-3\right)\cdot a\left(x\right)+4\)
Thay \(x=3\Leftrightarrow18+3a+a=4\Leftrightarrow4a=-14\Leftrightarrow a=-\dfrac{7}{2}\)
\(4,\Leftrightarrow ax^5+4x^4-9=\left(x-1\right)\cdot b\left(x\right)\)
Thay \(x=1\Leftrightarrow a+4-9=0\Leftrightarrow a=5\)
