Đáp án:$c.(3.7-5\dfrac{3}{11})-(10.7+2\dfrac{8}{11})=-15$
$d.\dfrac{5}{11}\cdot3\dfrac{7}{13}-\dfrac{5}{11}\cdot4\dfrac{7}{13}-\dfrac{6}{11}=-1$
Giải thích các bước giải:
c.Ta có :
$(3.7-5\dfrac{3}{11})-(10.7+2\dfrac{8}{11})$
$=3.7-5\dfrac{3}{11}-10.7-2\dfrac{8}{11}$
$=(3.7-10.7)-(5\dfrac{3}{11}+2\dfrac{8}{11})$
$=-7-(5+\dfrac{3}{11}+2+\dfrac{8}{11})$
$=-7-((5+2)+(\dfrac{3}{11}+\dfrac{8}{11}))$
$=-7-(7+\dfrac{3+8}{11})$
$=-7-(7+\dfrac{11}{11})$
$=-7-(7+1)$
$=-7-8$
$=-15$
d.Ta có :
$\dfrac{5}{11}\cdot3\dfrac{7}{13}-\dfrac{5}{11}\cdot4\dfrac{7}{13}-\dfrac{6}{11}$
$=(\dfrac{5}{11}\cdot3\dfrac{7}{13}-\dfrac{5}{11}\cdot4\dfrac{7}{13})-\dfrac{6}{11}$
$=\dfrac{5}{11}\cdot(3\dfrac{7}{13}-4\dfrac{7}{13})-\dfrac{6}{11}$
$=\dfrac{5}{11}\cdot((3+\dfrac{7}{13})-(4+\dfrac{7}{13}))-\dfrac{6}{11}$
$=\dfrac{5}{11}\cdot(3+\dfrac{7}{13}-4-\dfrac{7}{13})-\dfrac{6}{11}$
$=\dfrac{5}{11}\cdot((3-4)+(\dfrac{7}{13}-\dfrac{7}{13}))-\dfrac{6}{11}$
$=\dfrac{5}{11}\cdot(-1+0)-\dfrac{6}{11}$
$=\dfrac{5}{11}\cdot(-1)-\dfrac{6}{11}$
$=-\dfrac{5}{11}-\dfrac{6}{11}$
$=-(\dfrac{5}{11}+\dfrac{6}{11})$
$=-\dfrac{5+6}{11}$
$=-\dfrac{11}{11}$
$=-1$
