Mi aiutate a risolvere?

Allora ti aiuto! Cominciamo dal numeratore, poi il denominatore.

Facciamo i calcoli del numeratore: troviamo il mcm delle frazioni;

(4/2 + 1/2) * (140/20 - 36/20 - 105/20) + (8/12 + 27/12) * (7/35 - 10/35) =

= (5/2) * (- 1/20) + (35/12) * (- 3 /35) =

= (-  1/8) + ( - 1/4) = 

= (- 1/8) + (- 2/8) = - 3/8;  (numeratore).

Calcoli del denominatore:

(20/5 - 2/5) * (216/36 - 76/36 - 135/36) + (6/3 - 1/3) * (3/15 + 5/15 + 1/15)=

= (+ 18/5) * (+ 5/36) + (+ 5/3) * (+ 9/15) =

= (+ 1/2)  + (+ 1/1) =

= + 1/2 + 2/2 = + 3/2,  denominatore;

la frazione finale, è una divisione tra frazioni,  diventa:

= - 3/8 / (+ 3/2) =  ;    si trasforma in moltiplicazione con la frazione  inversa;

= - 3/8 * (+ 2/3) = - 1/4.

@andreag     Ciao 

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$\small \dfrac{\left(2+\dfrac{1}{2}\right)·\left(7-\dfrac{9}{5}-\dfrac{21}{4}\right)+\left(\dfrac{2}{3}+\dfrac{9}{4}\right)·\left(\dfrac{1}{5}-\dfrac{2}{7}\right)}{\left(4-\dfrac{2}{5}\right)·\left(6-\dfrac{19}{9}-\dfrac{15}{4}\right)+\left(2-\dfrac{1}{3}\right)·\left(\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{1}{15}\right)} =$

$\small = \dfrac{\left(\dfrac{4+1}{2}\right)·\left(\dfrac{140-36-105}{20}\right)+\left(\dfrac{8+27}{12}\right)·\left(\dfrac{7-10}{35}\right)}{\left(\dfrac{20-2}{5}\right)·\left(\dfrac{216-76-135}{36}\right)+\left(\dfrac{6-1}{3}\right)·\left(\dfrac{3+5+1}{15}\right)} =$

$\small = \dfrac{\dfrac{\cancel5^1}{2}·-\dfrac{1}{\cancel{20}_4}+\dfrac{\cancel{35}^1}{\cancel{12}_4}·-\dfrac{\cancel3^1}{\cancel{35}_1}}{\dfrac{\cancel{18}^1}{\cancel5_1}·\dfrac{\cancel5^1}{\cancel{36}_2}+\dfrac{\cancel5^1}{\cancel3_1}·\dfrac{\cancel9^3}{\cancel{15}_3}} =$

$\small = \dfrac{\dfrac{1}{2}·-\dfrac{1}{4}+\dfrac{1}{4}·-\dfrac{1}{1}}{\dfrac{1}{1}·\dfrac{1}{2}+\dfrac{1}{1}·\dfrac{\cancel3^1}{\cancel3_1}} =$

$\small = \dfrac{-\dfrac{1}{8}+\left(-\dfrac{1}{4}\right)}{\dfrac{1}{2}+\dfrac{1}{1}} =$

$\small = \dfrac{-\dfrac{1}{8}-\dfrac{1}{4}}{\dfrac{1+2}{2}} =$

$\small = \dfrac{\dfrac{-1-2}{8}}{\dfrac{3}{2}} =$

$\small = \dfrac{-\dfrac{3}{8}}{\dfrac{3}{2}} =$

$\small = -\dfrac{\cancel3^1}{\cancel8_4}·\dfrac{\cancel2^1}{\cancel3_1} =$

$\small = -\dfrac{1}{4}·\dfrac{1}{1} =$

$\small = -\dfrac{1}{4}$

((2 + 1/2)·(7 - 9/5 - 21/4) + (2/3 + 9/4)·(1/5 - 2/7))/((4 - 2/5)·(6 - 19/9 - 15/4) + (2 - 1/3)·(1/5 + 1/3 + 1/15)) =

=(5/2·(- 1/20) + 35/12·(- 3/35))/(18/5·(5/36) + 5/3·(3/5))=

=(- 1/8 + (- 1/4))/(1/2 + 1)=

=(- 3/8)/(3/2) =

=- 1/4