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Vậy 3x4 + x3 + 6x – 5 = (x2+ 1)(3x2 + x – 3) + 5x - 2
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Bài 1 :
a)\(\left(x^2+6x+9\right):\left(3+x\right)=\left(x+3\right)^2:\left(x+3\right)=x+3\)
b)\(\left(27x^3+1\right):\left(9x^2-3x+1\right)=\left(3x+1\right)\left(9x^2-3x+1\right):\left(9x^2-3x+1\right)=3x+1\)
c)\(\left(x^2+16-8x\right):\left(4-x\right)=\left(4-x\right)^2:\left(4-x\right)=4-x\)
\(\dfrac{A}{B}=\dfrac{2x^4-10x^3+3x^2-3x+2}{2x^2+1}=\dfrac{2x^4+x^2-10x^3-5x+2x^2+1+2x+1}{2x^2+1}\)
\(=x^2-x+1+\dfrac{2x+1}{2x^2+1}\)
R=2x+1
\(A=\left(x^2-x+1\right)\left(2x^2+1\right)+2x+1\)
Bài 2:
\(=\dfrac{3x^4+3x^2+x^3+x-3x^2-3+5x-5}{x^2+1}\)
\(=3x^2+x-3+\dfrac{5x-5}{x^2+1}\)
Bài 3:
\(\dfrac{A}{B}=\dfrac{2x^3-x^2-x+1}{x^2-2x}\)
\(=\dfrac{2x^3-4x^2+3x^2-6x+5x+1}{x^2-2x}\)
\(=2x^2+3+\dfrac{5x+1}{x^2-2x}\)
=>\(2x^3-x^2-x+1=\left(x^2-2x\right)\left(2x^2+3\right)+5x+1\)
Câu 1:
\(x^3+a\cdot x^2+2x+b\)
\(=x^3+x^2+x+\left(a-1\right)\cdot x^2+\left(a-1\right)\cdot x+\left(a-1\right)\) +x(2-a)+b-a+1
\(=\left(x^2+x+1\right)\left(x+a-1\right)\) +x(2-a)+b-a+1
Để \(x^3+a\cdot x^2+2x+b\) ⋮\(x^2+x+1\) thì 2-a=0 và b-a+1=0
=>a=2 và b=a-1=2-1=1
Câu 2:
a: \(\frac{A}{B}=\frac{4x^3-3x^2+1}{x^2+2x-1}\)
\(=\frac{4x^3+8x^2-4x-11x^2-22x+11+26x-10}{x^2+2x-1}\)
\(=4x-11+\frac{26x-10}{x^2+2x-1}\)
=>Q=4x-11; R=26x-10
b: \(\frac{A}{B}=\frac{3x^4-5x^3+7x^2-4x+2}{x^2-x+1}\)
\(=\frac{3x^4-3x^3+3x^2-2x^3+2x^2-2x+2x^2-2x+2}{x^2-x+1}\)
\(=3x^2-2x+2\)
=>\(Q=3x^2-2x+2\)
a: \(\dfrac{A}{B}=\dfrac{2x^5+3x^3-5x^2-11x+8}{x^3-3x+1}\)
\(=\dfrac{2x^5-6x^3+2x^2+9x^3-27x+9-7x^2+16x-1}{x^3-3x+1}\)
\(=2x^2+9+\dfrac{-7x^2+16x-1}{x^3-3x+1}\)
=>\(Q=2x^2+9;R=-7x^2+16x-1\)
b: R=-7x^2+16x-1
=-7(x^2-16x+1/7)
=-7(x^2-16x+64-449/7)
=-7(x-8)^2+449<=449
Dấu = xảy ra khi x=8
3x4 + x3 + 6x - 5
= (x2+1)( 3x2 + x-3)+5x-2