Question

please help me to solve the given question.​

Answer 1

Step-by-step explanation:

We have,

$\frac{ {x}^{2} - ax}{b} + \frac{ {x}^{2} - bx}{a} + \frac{ {x}^{2} - 3ax - 3bx }{a + b} = 0$

Taking x common from each term,

$x( \frac{x - a}{b} + \frac{x - b}{a} + \frac{x - 3a - 3b}{a + b}) = 0$

$either \: \: x = 0 \: \: or \: \: \frac{x - a}{b} + \frac{x - b}{a} + \frac{x - 3(a + b)}{a + b} = 0$

Now, consider the second case:

$\frac{x}{b} - \frac{a}{b} + \frac{x}{a} - \frac{b}{a} + \frac{x}{a + b} - 3 = 0$

$\implies( \frac{x}{b} + \frac{x}{a} + \frac{x}{a + b} ) - ( \frac{a}{b} + \frac{b}{a} + 3) = 0$

$\implies \: x( \frac{a(a + b) +b(a + b) + ab }{ab(a + b)} ) - ( \frac{ {a}^{2} + {b}^{2} - 3ab}{ab} ) =$

$\implies x(\frac{ {a}^{2} + {b}^{2} + 3ab }{ab(a + b)} ) + (\frac{ {a}^{2} + {b}^{2} - 3ab }{ab}) = 0$

$\implies x(\frac{ {a}^{2} + {b}^{2} + 3ab }{ab(a + b)} ) = - (\frac{ {a}^{2} + {b}^{2} - 3ab }{ab})$

$\implies \: x = - \frac{(a + b)( {a}^{2} + {b}^{2} - 3ab) }{( {a}^{2} + {b}^{2} + 3ab)}$