HELP! Parellel lines and transversal

Mathematics

1 answer:

amm18124 years ago

4 0

$m\angle 7=180-m\angle1=180-140=40$

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❊ Simplify :

Nataliya [291]

To simplify the given expression .

$\red{\frak{Given}}\Bigg\{ \sf \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 }$

$\rule{200}4$

$\sf\longrightarrow \small \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 } \\\\\\\sf\longrightarrow \small \dfrac{ x-1}{x^2-x -2x +2} +\dfrac{ x-2}{x^2-3x-2x+6} +\dfrac{ x -5}{x^2-5x -3x + 15 } \\\\\\\sf\longrightarrow\small \dfrac{ x -1}{ x ( x - 1) -2(x-1) } +\dfrac{ x-2}{x ( x -3) -2( x -3)} +\dfrac{ x -5}{ x(x-5) -3( x -5) } \\\\\\\sf\longrightarrow \small \dfrac{ x -1}{ ( x-2) (x-1) } +\dfrac{ x-2}{( x -2)(x-3) } +\dfrac{ x -5}{ (x-3)(x-5) } \\\\\\\sf\longrightarrow\small \dfrac{ 1}{ x-2} +\dfrac{ 1}{ x -3} +\dfrac{1}{ x -3} \\\\\\\sf\longrightarrow \small \dfrac{1}{x-2} +\dfrac{2}{x-3} \\\\\\\sf\longrightarrow \small \dfrac{ x-3 +2(x-2)}{ ( x -3)(x-2) } \\\\\\\sf\longrightarrow \small \dfrac{ x - 3 +2x -4 }{ (x-3)(x-2) } \\\\\\\sf\longrightarrow \underset{\blue{\sf Required \ Answer }}{\underbrace{\boxed{\pink{\frak{ \dfrac{ 3x -7}{ ( x -2)(x-3) } }}}}}$