Bài 2: Thực hiện phép cộng:
2x+7/x-3+x-4/x-2
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Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
a: \(=\dfrac{4x-2+6x^2-6x+2x^2+1}{2x\left(2x-1\right)}=\dfrac{8x^2-2x-1}{2x\left(2x-1\right)}\)
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
1) \(-6x^4+4x^3-2x^2\)
2) \(=x^2+4x-21-x^2-4x+5=-16\)
3) \(=6x^2-4x-x^2-4x-4=5x^2-8x-4\)
4) \(=2x^3-4x^2-8x-3x^2+6x+12=2x^3-7x^2-2x+12\)
c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
Bài 1:
(x² - 8)(x³ + 2x + 4)
= x².x³ + x².2x + x².4 - 8.x³ - 8.2x - 8.4
= x⁵ + 2x³ + 4x² - 8x³ - 16x - 32
= x⁵ - 6x³ + 4x² - 16x - 32
Bài 2
a) A(x) = -5/3 x² + 3/4 x⁴ + 2x - 7/3 x² - 2 + 4x + 1/4 x⁴
= (3/4 x⁴ + 1/4 x⁴) + (-5/3 x² - 7/3 x²) + (2x + 4x) - 2
= x⁴ - 4x² + 6x - 2
b) Bậc của A(x) là 4
Hệ số cao nhất là 1
`a, a/(a-3) - 3/(a+3) = (a(a+3) - 3(a-3))/(a^2-9)`
`= (a^2+9)/(a^2-9)`
`b, 1/(2x) + 2/x^2 = x/(2x^2) + 4/(2x^2) = (x+4)/(2x^2)`
`c, 4/(x^2-1) - 2/(x^2+x) = (4x)/(x(x-1)(x+1)) - (2(x-1))/(x(x+1)(x-1))`
`= (2x+2)/(x(x-1)(x+1)`
`= 2/(x(x-1))`
Answer:
\(\frac{2x+7}{x-3}+\frac{x-4}{x-2}\)
\(=\frac{\left(2x+7\right)\left(x-2\right)+\left(x-4\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\frac{2x^2+3x-14+x^2-7x+12}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{3x^2-4x-2}{\left(x-3\right)\left(x+2\right)}\)