Question

Pe mulţimea numerelor reale se definește legea de compoziție asociativă $x \circ y=x+y+11 x y$.

5p a) Demonstrați că $x \circ y=11\left(x+\frac{1}{11}\right)\left(y+\frac{1}{11}\right)-\frac{1}{11}$, pentru orice numere reale $x$ şi $y$.

$5 p$ b) Determinați numerele reale $x$, pentru care $x \circ x=\frac{8}{11}$.

$5 p$ c) Calculați partea întreagă a numărului $a=\left(1-\frac{1}{11}\right) \circ\left(1-\frac{2}{11}\right) \circ\left(1-\frac{3}{11}\right) \circ\left(1-\frac{4}{11}\right)$.

Answer 1

$x \circ y=x+y+11 x y$

a)

$x+y+11 x y=x+11xy+y+\frac{1}{11} -\frac{1}{11} =11x(y+\frac{1}{11} )+(y+\frac{1}{11} )-\frac{1}{11} =11(y+\frac{1}{11} )(x+\frac{1}{11})-\frac{1}{11}$

b)

$x \circ x=x+x+11 x^2=\frac{8}{11} \\\\11(x+\frac{1}{11})^2-\frac{1}{11}=\frac{8}{11}\\\\11(x+\frac{1}{11})^2=\frac{9}{11}\\\\(x+\frac{1}{11})^2=\frac{9}{121}\\\\(x+\frac{1}{11})=\frac{3}{11}\\\\ x=\frac{2}{11}\\\\ sau\\\\(x+\frac{1}{11})=-\frac{3}{11} \\\\x=-\frac{4}{11}$

c)

$a=(\frac{10}{11} \circ \frac{9}{11} )\circ (\frac{8}{11} \circ \frac{7}{11} )\\\\a=(11\cdot \frac{10}{11} -\frac{1}{11})\circ (11\cdot \frac{9}{11}\cdot \frac{8}{11}-\frac{1}{11} )\\\\ a= (10-\frac{1}{11})\circ (\frac{72}{11}-\frac{1}{11})\\\\a=11\cdot 10\cdot \frac{72}{11}-\frac{1}{11}\\\\a=720-\frac{1}{11}$

[a]=719

Un alt exercitiu cu legi de compozitie gasesti aici: https://brainly.ro/tema/9882261

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