Question

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Answer 1

Explicación paso a paso:

La suma o resta de fracciones es una de las operaciones básicas que permite combinar dos o más fracciones

Primera Columna A +B + C:

$\frac{3}{4} + \frac{1}{2}+ \frac{2}{3}=\frac{(3)(3) +(1)(6) +(2)(4) }{12} =\frac{9+6+8}{12} =\frac{23}{12}$

$\frac{7}{8} + \frac{3}{5}+ \frac{1}{2}=\frac{(7)(5) +(3)(8) +(1)(20) }{40} =\frac{35+24+20}{40} =\frac{79}{40}$

$\frac{9}{3} + \frac{9}{4}+ \frac{1}{4}=\frac{(9)(4) +(9)(3) +(1)(3) }{12} =\frac{36+27+3}{12} =\frac{66}{12} =\frac{11}{2}$

$\frac{15}{4} + \frac{1}{3}+ \frac{2}{5}=\frac{(15)(15) +(1)(20) +(2)(12) }{60} =\frac{225+20+24}{60} =\frac{269}{60}$

Segunda Columna A - B:

$\frac{3}{4} - \frac{1}{2}=\frac{(3)(1) - (1)(2) }{4} =\frac{3-2}{4} =\frac{1}{4}$

$\frac{7}{8} - \frac{3}{5}=\frac{(7)(5) - (3)(8) }{40} =\frac{35-24}{40} =\frac{11}{40}$

$\frac{9}{3} - \frac{9}{4}=\frac{(9)(4) - (9)(3) }{12} =\frac{36-27}{12} =\frac{9}{12} =\frac{3}{4}$

$\frac{15}{4} - \frac{1}{3}=\frac{(15)(3) - (1)(4) }{12} =\frac{45-4}{12} =\frac{41}{12}$

Tercera columna A+B - C:

$\frac{3}{4} + \frac{1}{2}- \frac{2}{3}=\frac{(3)(3) +(1)(6) -(2)(4) }{12} =\frac{9+6-8}{12} =\frac{7}{12}$

$\frac{7}{8} + \frac{3}{5}- \frac{1}{2}=\frac{(7)(5) +(3)(8) -(1)(20) }{40} =\frac{35+24-20}{40} =\frac{39}{40}$

$\frac{9}{3} + \frac{9}{4}- \frac{1}{4}=\frac{(9)(4) +(9)(3) -(1)(3) }{12} =\frac{36+27-3}{12} =\frac{60}{12} =5$

$\frac{15}{4} + \frac{1}{3}- \frac{2}{5}=\frac{(15)(15) +(1)(20) -(2)(12) }{60} =\frac{225+20-24}{60} =\frac{221}{60}$