Đáp án: a.$4$ b.$2020$ c.$-\dfrac{13}{20}$
Giải thích các bước giải:
a.Ta có:
$\dfrac17.16\dfrac23+\dfrac{-1}{7}.44\dfrac23+(-2)^3$
$=\dfrac17.16\dfrac23-\dfrac{1}{7}.44\dfrac23+(-2)^3$
$=\dfrac17.(16\dfrac23-44\dfrac23)+(-2)^3$
$=\dfrac17.((16+\dfrac23)-(44+\dfrac23))+(-2)^3$
$=\dfrac17.(16+\dfrac23-44-\dfrac23)+8$
$=\dfrac17.(16-44+\dfrac23-\dfrac23)+8$
$=\dfrac17.(-28+0)+8$
$=\dfrac17.(-28)+8$
$=-4+8$
$=4$
b.Ta có:
$\dfrac{(-9)^2.3^3}{3^7}.2020$
$=\dfrac{9^2.3^3}{3^7}.2020$
$=\dfrac{(3^2)^2.3^3}{3^7}.2020$
$=\dfrac{3^4.3^3}{3^7}.2020$
$=\dfrac{3^{4+3}}{3^7}.2020$
$=\dfrac{3^{7}}{3^7}.2020$
$=1.2020$
$=2020$
c.Ta có:
$\dfrac{(-1)^3}{15}+(-\dfrac23)^2:2\dfrac23-|-\dfrac56|$
$=\dfrac{-1}{15}+\dfrac23:(2+\dfrac23)-\dfrac56$
$=\dfrac23:\dfrac83-\dfrac56+\dfrac{-1}{15}$
$=\dfrac14-(\dfrac56+\dfrac{1}{15})$
$=\dfrac14-(\dfrac56+\dfrac{1}{15})$
$=\dfrac14-\dfrac9{10}$
$=-\dfrac{13}{20}$
