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Răspuns:

8

Explicație pas cu pas:

[1 1/2+1,1(6)+1,(3)]×[(1/3)²+(3/2)⁵:(3/2)³-13/36]

introducem intregii in fractie

1 1/2=3/2

transformăm fractiile zecimale periodice in fractii ordinare

1,1(6)=(116-11)/90=105/90 simplificam cu 15=7/6

iar 1,(3)=(13-1)/9=12/9 simplificam cu 3=4/3

deci ne rămâne...

(3/2+7/6+4/3)×[(1/3)²+(3/2)⁵:(3/2)³-13/36]

calculam suma

4×[(1/3)²+(3/2)⁵:(3/2)³-13/36]

avem (1/3)²=1²/3²=1/9

4×[1/9+(3/2)⁵:(3/2)³-13/36]

calculam catul

4×[1/9+(3/2)²-13/36]

Avem (3/2)²=3²/2²=9/4

4×(1/9+9/4-13/36)

calculam suma si diferenta

4×2

înmulțim numerele

8

[tex]\it [1\dfrac{1}{2}+1,1(6)+1,(3)]\cdot[(\dfrac{1}{3})^2+(\dfrac{3}{2})^5:(\dfrac{3}{2})^3-\dfrac{13}{36}] =\\ \\ \\ =\Big(\dfrac{3}{2}+1\dfrac{15^{(15}}{90}+1\dfrac{3^{(3}}{9}\Big)\cdot\Big(\dfrac{^{4)}1}{9}+\dfrac{^{9)}9}{4}-\dfrac{13}{36}\Big)= \Big(\dfrac{^{3)}3}{2}+\dfrac{7}{6}+\dfrac{^{2)}4}{3}\Big)\cdot\dfrac{4+81-13}{36}=\\ \\ \\= \dfrac{24}{6}\cdot\dfrac{72}{36}=4\cdot2= 8[/tex]