6x/x mũ 2-9+5x/x-3+x/x+3
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Bài 1:
1; 5\(x\) + \(x\) = 39 - 311 : 39
\(x\).(5 + 1) = 39 - 32
\(x.6\) = 39 - 9
\(x.6\) = 30
\(x\) = 30 : 6
\(x\) = 5
Vậy \(x\) = 5
2; 5\(x\) + \(x\) = 150 : 2 + 3
\(x\).(5 + 1) = 75 + 3
\(x.6\) = 78
\(x\) = 78 : 6
\(x\) = 13
Vậy \(x=13\)
d: \(\left(6x-5x^2-15+2x^3\right):\left(2x-5\right)\)
\(=\frac{2x^3-5x^2+6x-15}{2x-5}\)
\(=\frac{x^2\left(2x-5\right)+3\left(2x-5\right)}{2x-5}=x^2+3\)
e: \(\left(x^3+x^5+x^2+1\right):\left(x^3+1\right)\)
\(=\frac{x^5+x^3+x^2+1}{x^3+1}=\frac{x^2\left(x^3+1\right)+\left(x^3+1\right)}{x^3+1}\)
\(=x^2+1\)
i: \(\left(3-2x+2x^3+5x^2\right):\left(2x^2-x+1\right)\)
\(=\frac{2x^3+5x^2-2x+3}{2x^2-x+1}=\frac{2x^3-x^2+x+6x^2-3x+3}{2x^2-x+1}\)
\(=\frac{x\left(2x^2-x+1\right)+3\left(2x^2-x+1\right)}{2x^2-x+1}=x+3\)
\(a.\frac{x^3+6x^2+2x-3}{x^2+5x-3}=\frac{\left(x+1\right)\left(x^2+5x-3\right)}{x^2+5x-3}=x+1\)
\(b.\frac{x^3-3x^2+x-3}{x-3}=\frac{\left(x-3\right)\left(x^2+1\right)}{x-3}=x^2+1\)
\(c.\frac{x^2+3x-10}{x-2}=\frac{\left(x-2\right)\left(x+5\right)}{x-2}=x+5\)

\(=\dfrac{6x}{\left(x-3\right)\left(x+3\right)}-\dfrac{9+5x}{x-3}+\dfrac{x}{x+3}\)
\(=\dfrac{6x-\left(5x+9\right)\left(x+3\right)+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+3x-5x^2-15x-9x-27}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-4x^2-21x-27}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-\left(4x^2+12x+9x+27\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-4x-9}{x-3}\)