Question

(1/6-1/5)*(4-9/4)+3*(-1/9)

(4/3+ 9/2) – (3/10+ 2/5* 1/4) * 7/8 =

– (6/5+2* 8/3) – (- 1/4-3) : 5/2 =

[9/13- 4/11* 2/3+ 17/2 : (-1/6)] + 2 =

(1/4 * ((3 )/5-( 5)/6))/(1/2 - 3∶ 2/3)

Answer 1

a) (1/6-1/5)*(4-9/4)+3*(-1/9)

=       b) (4/3+ 9/2) – (3/10+ 2/5* 1/4) * 7/8 =  

c) – (6/5+2* 8/3) – (- 1/4-3) : 5/2 =      d) [9/13- 4/11* 2/3+ 17/2 : (-1/6)] + 2 =  

e) (1/4 * ((3 )/5-( 5)/6))/(1/2 - 3∶ 2/3)

=

Hola!!!!

Primero se resuelven los paréntesis y luego los corchetes, En cada paréntesis el orden a seguir es primero las multiplicaciones y divisiones; después las sumas y restas. Entonces

$b)\left(\dfrac{4}{3}+\dfrac{9}{2}\right)-\left(\dfrac{3}{10}+\dfrac{2}{5}*\dfrac{1}{4}\right)*\dfrac{7}{8}=\\\\\\\left(\dfrac{8}{6}+\dfrac{27}{6}\right)-\left(\dfrac{3}{10}+\dfrac{2*1}{5*4}\right)*\dfrac{7}{8}=\\\\\\\left(\dfrac{8+27}{6}\right)-\left(\dfrac{3}{10}+\dfrac{2}{20}\right)*\dfrac{7}{8}=\\\\\\\left(\dfrac{5}{30}-\dfrac{6}{30}\right)*\left(\dfrac{36}{9}-\dfrac{4}{9}\right)+\left(-\dfrac{3*1}{9}\right)=\\\\\\\left(-\dfrac{1}{30}\right)*\left(\dfrac{32}{9}\right)+\left(-\dfrac{3}{9}\right)=\\\\\\\left(-\dfrac{32}{270}\right)+\left(-\dfrac{3}{9}\right)=\qquad\qquad simplificamos\\\\\\\left(-\dfrac{16}{135}\right)+\left(-\dfrac{1}{3}\right)=$

$\left(\dfrac{35}{6}\right)-\left(\dfrac{6}{20}+\dfrac{2}{20}\right)*\dfrac{7}{8}=\\\\\\\left(\dfrac{35}{6}\right)-\left(\dfrac{8}{20}\right)*\dfrac{7}{8}=\qquad primero \ la \ multiplicacion\\\\\\\dfrac{35}{6}-\dfrac{8*7}{20*8}=\dfrac{35}{6}-\dfrac{7}{20}=\dfrac{350}{60}-\dfrac{21}{60}=\boxed{\frac{329}{60}}$

$c)-\left(\dfrac{6}{5}+2*\dfrac{8}{3}\right)-\left(-\dfrac{1}{4}-3\right):\dfrac{5}{2}=\\\\\\-\left(\dfrac{6}{5}+\dfrac{16}{3}\right)-\left(-\dfrac{1}{4}-\dfrac{12}{4}\right):\dfrac{5}{2}=\\\\\\-\left(\dfrac{18}{15}+\dfrac{80}{15}\right)-\left(-\dfrac{13}{4}\right):\dfrac{5}{2}=\\\\\\-\left(\dfrac{98}{15}\right)+\dfrac{13}{4}:\dfrac{5}{2}=\\\\\\-\dfrac{98}{15}+\dfrac{13}{4}\right):\dfrac{5}{2}=-\dfrac{98}{15}+\dfrac{26}{20}=\dfrac{-392+78}{60}=-\dfrac{314}{60}=\boxed{-\frac{157}{30}}$

$d) \left[\dfrac{9}{13}-\dfrac{4}{11}*\dfrac{2}{3}+\dfrac{17}{2}:\left(-\dfrac{1}{6}\right)\right]+2=\\\\\\\left[\dfrac{9}{13}-\dfrac{4*2}{11*3}+\dfrac{17*6}{2*(-1)}\right]+2=\\\\\\\left[\dfrac{9}{13}-\dfrac{8}{33}+\dfrac{102}{(-2)}\right]+2=\\\\\\\left[\dfrac{9}{13}-\dfrac{8}{33}-51}\right]+2=\\\\\\\left[\dfrac{297}{429}-\dfrac{104}{429}-\dfrac{21879}{429} 51}\right]+2=-\dfrac{21686}{429}+2=\dfrac{-21686+858}{429}=\boxed{-\frac{20828}{429}}$

$e) \left[\dfrac{1}{4}*\left(\dfrac{3}{5}-\dfrac{5}{6}\right):\left(\dfrac{1}{2}-3:\dfrac{2}{3}\right)\right]=\\\\\\\ \left[\dfrac{1}{4}*\left(\dfrac{18}{30}-\dfrac{25}{30}\right):\left(\dfrac{1}{2}-\dfrac{9}{2}\right)\right]\\\\\\\ \left[\dfrac{1}{4}*\left(-\dfrac{7}{30}\right):\left(-\dfrac{8}{2}\right)\right]=\dfrac{1}{4}*\dfrac{7}{120}=\boxed{\dfrac{7}{480}}$

Espero que te sirva, salu2!!!!