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\(\left[\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2\right]:\left(x^2-6x+9\right)=\left[\left(3-x\right)^5-7\left(3-x\right)^4-4\left(3-x\right)^2\right]:\left(3-x\right)^2=\left(3-x\right)^2\left[\left(3-x\right)^3-7\left(3-x\right)^2-4\right]:\left(3-x\right)^2=\left(3-x\right)^3-7\left(3-x\right)^2-4=27-27x+9x^2-x^3-63+42x-7x^2-4=-x^3+2x^2+15x-40\)
\(\dfrac{\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{x^2-6x+9}\)
\(=\dfrac{-\left(x-3\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{\left(x-3\right)^2}\)
\(=-\left(x-3\right)^3-7\left(x-3\right)^2-4\)
\(=\left(3x^4-3x^3+x^3-x^2+8x^2-8x+9x-9\right):\left(x-1\right)\\ =\left(x-1\right)\left(3x^3+x^2+8x+9\right):\left(x-1\right)\\ =3x^3+x^2+8x+9\)
2)a)3x(x-5)-(x-1)(2+3x)=30
<=>3x2-15x-3x2+x+2=30
<=>-14x+2=30
<=>-14x=30-2
<=>-14x=28
<=>x=-2
b)(x+2)(x+3)-(x-2)(x+5)=0
<=>x2+5x+6-x2-3x+10=0
<=>2x+16=0
<=>2x=-16
<=>x=-8
c)(3x+2)(2x+9)-(x+2)(6x+1)=9
<=>6x2+31x+18-6x2-13x-2=9
<=>18x+16=9
<=>18x=9-16
<=>18x=-7
<=>x=-7/18
\(\left(x^3-x^2-5x-3\right):\left(x-3\right)\\ =\left[\left(x^3-3x^2\right)+\left(2x^2-6x\right)+\left(x-3\right)\right]:\left(x-3\right)\\ =\left[x^2\left(x-3\right)+2x\left(x-3\right)+\left(x-3\right)\right]:\left(x-3\right)\\ =\left[\left(x-3\right)\left(x^2+2x+1\right)\right]:\left(x-3\right)\\ =x^2+2x+1\)
a, \(\left(x^2-9\right)^2-\left(x-3\right)\left(x+3\right)\left(x^2+9\right)=\left(x^2-9\right)^2-\left(x^2-9\right)\left(x^2+9\right)\)
\(=x^4-18x^2+81-x^4+81=-18x^2+162\)
b, \(\left(x^2+x-3\right)\left(x^2-x+3\right)=\left[x^4-\left(x-3\right)^2\right]\)
\(=x^4-x^2+6x-9\)
Ta có phép chia