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b: \(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
b: \(\dfrac{A\left(x\right)}{B\left(x\right)}=\dfrac{x^4-\dfrac{1}{2}x^3+\dfrac{1}{2}x^3-\dfrac{1}{4}x^2+\dfrac{9}{4}x^2-\dfrac{9}{8}x-\dfrac{15}{8}x+\dfrac{15}{16}+a-\dfrac{1}{16}}{2x-1}\)
Để A(x) chia hết cho B(x) thì a-1/16=0
hay a=1/16
a: \(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
1) \(\Leftrightarrow\left(x-4\right)\left(x+4\right)-x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4-x\right)=0\)
\(\Leftrightarrow\left(x-4\right)4=0\)
\(\Leftrightarrow x=4\)
2) \(\left(x+3\right)^2-\left(x-3\right)\left(x+5\right)=x^2+6x+9-x^2-2x+15=4x+24\)
3) \(2x^3+3x^2-2x+a=2x^2\left(x-2\right)+7x\left(x-2\right)+16\left(x-2\right)+32+a\)
Để \(2x^3+3x^2-2x+a⋮x-2\) thì \(32+a=0\Leftrightarrow a=-32\)
1.
x2 - 16 - x(x - 4) = 0
<=> (x2 - 42) - x(x - 4) = 0
<=> (x - 4)(x + 4) - x(x - 4) = 0
<=> (x + 4 - x)(x + 4) = 0
<=> 4(x + 4) = 0
<=> x + 4 = 0
<=> x = -4
2.
(x + 3)2 - (x - 3)(x + 5)
= x2 + 6x + 9 - (x2 + 5x - 3x - 15)
= x2 + 6x + 9 - x2 + 5x - 3x - 15
= x2 - x2 + 6x + 5x - 3x + 9 - 15
= 8x - 6
a) \(A\left(x\right)=2x^3-x^2-x+1\)
\(=\left(2x^3-4x^2\right)+\left(3x^2-6x\right)+\left(5x-10\right)+11\)
\(=\left(x-2\right).\left(2x^2+3x+5\right)+11\)
Vậy \(A\left(x\right):B\left(x\right)=2x^2+3x+5\) dư \(11\)
b) Để \(A\left(x\right)⋮B\left(x\right)\) thì \(11⋮B\left(x\right)\)
\(\Rightarrow x-2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
\(\Rightarrow x\inơ\left\{13;3;2;-9\right\}\)
