Xfhjjkkkjjhhhhhhhfdsss Question Xfhjjkkkjjhhhhhhhfdsss in progress 0 Môn Toán Kiệt Gia 6 years 2020-11-28T00:46:00+00:00 2020-11-28T00:46:00+00:00 2 Answers 60 views 0
Answers ( )
Đáp án:
$1)
a)\dfrac{7}{6}\\
b)1\\
c)-4\\
2)
a)x=\dfrac{-2}{3}\\
b)
{\left[\begin{aligned}x=\dfrac{3}{7}\\x=\dfrac{1}{7}\end{aligned}\right.}\\
c)
{\left[\begin{aligned}x= \dfrac{33}{40} \\x=-\dfrac{23}{40} \end{aligned}\right.}\\$
Giải thích các bước giải:
$1)
a) \dfrac{5}{3}+\dfrac{2}{3}.\left ( \dfrac{-3}{4} \right )\\
=\dfrac{5}{3}+\dfrac{2}{3}.\dfrac{-3}{4}\\
=\dfrac{5}{3}-\dfrac{1}{2}\\
=\dfrac{10}{6}-\dfrac{3}{6}\\
=\dfrac{7}{6}\\
b) \dfrac{-2}{7}.\dfrac{8}{17}-\dfrac{2}{7}.\dfrac{9}{17}+1\dfrac{2}{7}\\
=\dfrac{2}{7}.\left ( \dfrac{-8}{17}-\dfrac{9}{17} \right )+\dfrac{9}{7}\\
=\dfrac{2}{7}.\left ( \dfrac{-17}{17}+\dfrac{9}{2} \right )\\
=\dfrac{2}{7}.\left ( \dfrac{-2}{2}+\dfrac{9}{2} \right )\\
=\dfrac{2}{7}.\dfrac{7}{2}\\
=1\\
c) (-2)^3+3\sqrt{\dfrac{16}{9}}+\left | -5\dfrac{1}{3} \right |+\left ( \dfrac{-2017}{2018} \right )^0:\dfrac{-3}{16}\\
=-8+3.\dfrac{4}{3}+5\dfrac{1}{3}+1.\dfrac{-16}{3}\\
=-8+4+\dfrac{16}{3}-\dfrac{16}{3}\\
=-4\\
2)
a) \dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\\
\Leftrightarrow \dfrac{2}{5}x=\dfrac{29}{60}-\dfrac{3}{4}\\
\Leftrightarrow \dfrac{2}{5}x=\dfrac{29}{60}-\dfrac{45}{60}\\
\Leftrightarrow \dfrac{2}{5}x=\dfrac{-16}{60}=\dfrac{-4}{15}\\
\Leftrightarrow x=\dfrac{-4}{15}.\dfrac{5}{2}=\dfrac{-2}{3}\\
b)
\left | x-\dfrac{2}{7} \right |=\dfrac{-1}{5}.\dfrac{-5}{7}\\
\Leftrightarrow \left | x-\dfrac{2}{7} \right |=\dfrac{1}{7}\\
\Leftrightarrow {\left[\begin{aligned}x-\dfrac{2}{7}=\dfrac{1}{7}\\x-\dfrac{2}{7}=\dfrac{-1}{7}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{1}{7}+\dfrac{2}{7}\\x=\dfrac{-1}{7}+\dfrac{2}{7}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{3}{7}\\x=\dfrac{1}{7}\end{aligned}\right.}\\
c)
\left ( 2x-\dfrac{1}{4} \right )^2=\dfrac{49}{25}\\
\Leftrightarrow 2x-\dfrac{1}{4} =\pm \dfrac{7}{5}\\
\Leftrightarrow {\left[\begin{aligned}2x-\dfrac{1}{4} = \dfrac{7}{5}\\2x-\dfrac{1}{4} =-\dfrac{7}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x= \dfrac{7}{5}+\dfrac{1}{4} \\2x=-\dfrac{7}{5}+\dfrac{1}{4} \end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x= \dfrac{28}{20}+\dfrac{5}{20} \\2x=-\dfrac{28}{20}+\dfrac{5}{20} \end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x= \dfrac{33}{20} \\2x=-\dfrac{23}{20} \end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x= \dfrac{33}{40} \\x=-\dfrac{23}{40} \end{aligned}\right.}\\$
Đáp án:
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