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Answer 1

Required Answer:-

Given:

• $\sf \dfrac{1}{x - 1} - \dfrac{1}{x} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}$

To Find:

• The value of x.

Solution:

Given that,

$\sf \implies \dfrac{1}{x - 1} - \dfrac{1}{x} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}$

→ LCM of (x - 1) and x is - x(x - 1)

$\sf \implies \dfrac{x - (x - 1)}{x(x - 1)} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}$

$\sf \implies \dfrac{x - x + 1}{x(x - 1)} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}$

$\sf \implies \dfrac{1}{ {x}^{2} - x} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}$

→ Again, LCM of (x + 3) and (x + 4) is - (x + 3)(x + 4)

$\sf \implies \dfrac{1}{ {x}^{2} - x} = \dfrac{(x + 4) - (x + 3)}{(x + 3)(x + 4)}$

$\sf \implies \dfrac{1}{ {x}^{2} - x} = \dfrac{x + 4 -x - 3}{(x + 3)(x + 4)}$

$\sf \implies \dfrac{1}{ {x}^{2} - x} = \dfrac{1}{(x + 3)(x + 4)}$

On cross multiplying, we get,

$\sf \implies (x + 3)(x + 4)= {x}^{2} - x$

$\sf \implies x(x + 4) + 3(x + 4)= {x}^{2} - x$

$\sf \implies {x}^{2} + 4x + 3x +12={x}^{2} - x$

$\sf \implies {x}^{2} + 7x +12={x}^{2} - x$

$\sf \implies {x}^{2} + 7x +12 - {x}^{2} + x = 0$

$\sf \implies 8x +12 = 0$

$\sf \implies 4(2x + 3) = 0$

$\sf \implies 2x + 3 = \dfrac{0}{4}$

$\sf \implies 2x + 3 = 0$

$\sf \implies 2x = - 3$

$\sf \implies x = \dfrac{ - 3}{2}$

→ Hence, the value of x is -3/2.

Answer:

• x = -3/2.

•••♪

Answer 2

Answer:

• x = -1.5

Given:

$: \implies \sf \dfrac{1}{x-1} - \dfrac{1}{x} = \dfrac{1}{x+3} - \dfrac{1}{x+4}$

To Find:

The value of x.

Solution:

$: \implies \sf \dfrac{1}{x-1} - \dfrac{1}{x} - \dfrac{1}{x+3} + \dfrac{1}{x+4} = 0$

$: \implies \sf \dfrac{3x^2 + 12x - x^3 - 6x^2 + 12 + 4x + x^3 + 3x^2 - 3x}{x \times (x-1) \times (x+3) \times (x+4)}$

$: \implies \sf \dfrac{0+8x+12}{x \times (x-1) \times (x+3) \times (x+4)} = 0$

$: \implies \sf \dfrac{8x + 12}{x \times (x-1) \times (x+3) \times (x+4)} = 0$

$: \implies \sf 8x + 12 = 0$

$: \implies \sf 8x = -12$

$: \implies \sf x = - \dfrac{3}{2}$

$: \implies \sf x = -1.5$

$\large \dag {\underline{\boxed{\sf x = -1.5}}}$