Đáp án:
$1)
a)\dfrac{1}{5}\\
b)\dfrac{-1}{2}\\
c)\dfrac{2}{7}\\
d)
\dfrac{147}{200}\\
2)
a) x=4\\
b)
x=-1\\
c)x=1\\
d) {\left[\begin{aligned}x=\dfrac{7}{6}\\x=\dfrac{1}{6}\end{aligned}\right.}\\$
Giải thích các bước giải:
$1)
a)\dfrac{3}{5}+\dfrac{-2}{5}=\dfrac{3-2}{5}=\dfrac{1}{5}\\
b)\left ( \dfrac{4}{5}+\dfrac{1}{2} \right ).\left ( \dfrac{3}{13}-\dfrac{8}{13} \right )\\
=\left ( \dfrac{8}{10}+\dfrac{5}{10} \right ).\dfrac{-5}{13}\\
=\dfrac{13}{10}.\dfrac{-5}{13}\\
=\dfrac{-1}{2}\\
c)\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\\
=\dfrac{-5}{7}.\left ( \dfrac{2}{11}+\dfrac{9}{11} \right )+1\\
=\dfrac{-5}{7}.\dfrac{11}{11}+\dfrac{7}{7}\\
=\dfrac{2}{7}\\
d)
\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+...+\dfrac{3}{98.100}\\
=\dfrac{3}{2}.\left (\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100} \right )\\
=\dfrac{3}{2}.\left (\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100} \right )\\
=\dfrac{3}{2}.\left ( \dfrac{1}{2}-\dfrac{1}{100} \right )\\
=\dfrac{3}{2}.\left ( \dfrac{50}{100}-\dfrac{1}{100} \right )\\
=\dfrac{3}{2}.\dfrac{49}{100}\\
=\dfrac{147}{200}\\
2)
a) 2x+8=16\\
\Leftrightarrow 2x=16-8=8\\
\Leftrightarrow x=\dfrac{8}{2}=4\\
b)
x-\dfrac{-5}{12}=\dfrac{-7}{12}\\
\Leftrightarrow x=\dfrac{-7}{12}+\dfrac{-5}{12}\\
\Leftrightarrow x=\dfrac{-12}{12}=-1\\
c)\dfrac{x}{20}=\dfrac{7}{10}+\dfrac{-13}{20}\\
\Leftrightarrow \dfrac{x}{20}=\dfrac{14}{20}+\dfrac{-13}{20}\\
\Leftrightarrow \dfrac{x}{20}=\dfrac{1}{20}\\
\Leftrightarrow x=\dfrac{1}{20}.20=1\\
d) \left | x-\dfrac{2}{3} \right |+0,25=\dfrac{3}{4}\\
\Leftrightarrow \left | x-\dfrac{2}{3} \right |=\dfrac{3}{4}-\dfrac{1}{4}\\
\Leftrightarrow \left | x-\dfrac{2}{3} \right |=\dfrac{1}{2}\\
\Leftrightarrow {\left[\begin{aligned}x-\dfrac{2}{3}=\dfrac{1}{2}\\x-\dfrac{2}{3}=-\dfrac{1}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{1}{2}+\dfrac{2}{3}\\x=-\dfrac{1}{2}+\dfrac{2}{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{3}{6}+\dfrac{4}{6}\\x=-\dfrac{3}{6}+\dfrac{4}{6}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{7}{6}\\x=\dfrac{1}{6}\end{aligned}\right.}\\$
