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a: \(\left(19x^2-14x^3+9-20x+2x^4\right):\left(1+x^2-4x\right)\)
\(=\left(2x^4-14x^3+19x^2-20x+9\right):\left(x^2-4x+1\right)\)
\(=\left(2x^4-8x^3+2x^2-6x^3+24x^2-6x-7x^2+28x-7-42x+16\right)\) :\(\left(x^2-4x+1\right)\)
\(=2x^2-6x-7+\frac{-42x+16}{x^2-4x+1}\)
b: \(\left(3x^4-2x^3-2x^2+4x+8\right):\left(x^2-2\right)\)
\(=\left(3x^4-6x^2-2x^3+4x+4x^2-8+16\right):\left(x^2-2\right)\)
\(=3x^2-2x+4+\frac{16}{x^2-2}\)
c: \(\left(2x^3-26x-24\right):\left(x^2+4x+3\right)\)
\(=\left(2x^3+8x^2+6x-8x^2-32x-24\right):\left(x^2+4x+3\right)\)
=2x-8
\(a,2x^3+5x^2+5x+3\)
\(=2x^3+3x^2+2x^2+3x+2x+3\)
\(=x^2\left(2x+3\right)+x\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(2x+3\right)\left(x^2+x+1\right)\)
a) 4x2 - 2x + 3 - 4x.(x - 5) = 7x - 3
--> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
--> 4x2 - 2x - 4x2 + 20x - 7x = -3 - 3
--> 11x = -6
--> x = \(\frac{-6}{11}\)
b) -3x.(x - 5) + 5.(x - 1) + 3x2 = 4x
--> -3x2 + 15x + 5x - 5 + 3x2 = 4x
--> -3x2 + 15x + 5x + 3x2 - 4x = 5
--> 16x = 5
--> x = \(\frac{5}{16}\)
c) 7x.(x - 2) - 5.(x - 1) = 21x2 - 14x2 + 3
--> 7x2 - 14x - 5x + 5 = 7x2 + 3
--> 7x2 - 14x - 5x - 7x2 = -5 + 3
--> -19x = -2
--> x = \(\frac{2}{19}\)
d) 3.(5x - 1) - x.(x - 2) + x2 - 13x = 7
--> 15x - 3 - x2 + 2x + x2 - 13x = 7
--> 15x - x2 + 2x + x2 - 13x = 3 + 7
--> 4x = 10
--> x = \(\frac{5}{2}\)
e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12
--> 2x2 - 3x - 2x2 + 10x = 12
--> 7x = 12
--> x = \(\frac{12}{7}\)
~ Học tốt ~
a) 4x2 - 2x + 3 - 4x(x - 5) = 7x - 3
=> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
=> 18x + 3 = 7x - 3
=> 18x - 7x = -3 - 3
=> 11x = -6
=> x = -6/11
b) -3x(x - 5) + 5(x - 1) + 3x2 = 4x
=> -3x2 + 15x + 5x - 5 + 3x2 = 4x
=> 20x - 5 = 4x
=> 20x - 4x = 5
=> 16x = 5
=> x = 5/16
\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)
\(\Leftrightarrow7x^2-7x^2-19x=3-5\)
\(\Leftrightarrow-19x=-2\)
\(\Leftrightarrow x=\frac{2}{19}\)
a: \(x^3-x^2-14x+24\)
\(=x^3+x-12-x^2-15x+36\)
=>\(\left(x^3-x^2-14x+24\right):\left(x^3+x-12\right)=1+\frac{-x^2-15x+36}{x^3+x-12}\)
Để dư là 0 thì \(-x^2-15x+36=0\)
=>\(x^2+15x-36=0\) (1)
\(\Delta=15^2-4\cdot1\cdot\left(-36\right)=225+144=369>0\)
Do đó: (1) có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}x=\frac{-15-\sqrt{369}}{2\cdot1}=\frac{-15-3\sqrt{41}}{2}\\ x=\frac{-15+3\sqrt{41}}{2}\end{array}\right.\)
b: \(x^5+4x^3+3x^2-5x+15\)
\(=x^5-x^3+3x^2+5x^3-5x+15=\left(x^3-x+3\right)\left(x^2+5\right)\)
=>\(\frac{x^5+4x^3+3x^2-5x+15}{x^3-x+3}=x^2+5\)
=>Đây là phép chia hết
c: \(2x^4+2x^3+3x^2-5x-20\)
\(=2x^4+2x^3+8x^2-5x^2-5x-20=\left(x^2+x+4\right)\left(2x^2-5\right)\)
=>\(\frac{2x^4+2x^3+3x^2-5x-20}{x^2+x+4}=2x^2-5\)
d: \(2x^4-14x^3+19x^2-20x+9\)
\(=2x^4-8x^3+2x^2-6x^3+24x^2-6x-7x^2+28x-7-42x+16\)
\(=\left(x^2-4x+1\right)\left(2x^2-6x-7\right)-42x+16\)
=>\(\frac{2x^4-14x^3+19x^2-20x+9}{x^2-4x+1}=2x^2-6x-7\) dư -42x+16
để dư bằng 0 thì -42x+16=0
=>-42x=-16
=>\(x=\frac{16}{42}=\frac{8}{21}\)