Đáp án:
Giải thích các bước giải:
$b,-3\dfrac{1}{7}.\dfrac{12}{35}+\dfrac{-5}{6}.\dfrac{-12}{35}-1\dfrac{1}{7}.\dfrac{12}{-35}$
$=\dfrac{-22}{7}.\dfrac{12}{35}+\dfrac{5}{6}.\dfrac{12}{35}+\dfrac{8}{7}.\dfrac{12}{35}$
$=\left (\dfrac{-22}{7}+\dfrac{8}{7}+\dfrac{5}{6} \right ).\dfrac{12}{35}$
$=\left (\dfrac{5}{6}-2 \right ).\dfrac{12}{35}$
$=\dfrac{-7}{6}.\dfrac{12}{35}$
$=-\dfrac{2}{5}$
$d,-4\dfrac{5}{9}.\dfrac{15}{16}-\dfrac{15}{16}.2\dfrac{4}{9}+2\dfrac{1}{5}.\dfrac{15}{16}$
$=\left (\dfrac{-41}{9}-\dfrac{22}{9}+\dfrac{11}{5} \right ).\dfrac{15}{16}$
$=\left (\dfrac{11}{5}-7 \right ).\dfrac{15}{16}$
$=\dfrac{-24}{5}.\dfrac{15}{16}$
$=-\dfrac{9}{2}$
Bài 2:
$a,-\dfrac{5}{6}+\left |x-\dfrac{2}{3} \right |=0,25$
$⇔\left |x-\dfrac{2}{3} \right |=\dfrac{1}{4}+\dfrac{5}{6}$
$⇔\left |x-\dfrac{2}{3} \right |=\dfrac{13}{12}$
⇔\(\left[ \begin{array}{l} x-\dfrac{2}{3}=\dfrac{13}{12} \\x-\dfrac{2}{3}=-\dfrac{13}{12}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{7}{4}\\x=-\dfrac{5}{12}\end{array} \right.\)
$3-\left |\dfrac{3}{4}-x \right |=-\dfrac{1}{2}$
$⇔\left |\dfrac{3}{4}-x \right |=3+\dfrac{1}{2}$
$⇔\left |\dfrac{3}{4}-x \right |=\dfrac{7}{2}$
⇔\(\left[ \begin{array}{l}\dfrac{3}{4}-x=\dfrac{7}{2}\\\dfrac{3}{4}-x=\dfrac{-7}{2}\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\dfrac{-11}{4}\\x=\dfrac{17}{4}\end{array} \right.\)
Chúc em học tốt.
