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a) \(\frac{3}{4}\)+ \(\frac{4}{3}\):\(\frac{2}{7}\)- \(\frac{4}{3}\)
=\(\frac{3}{4}\) +\(\frac{4}{3}\) .\(\frac{7}{2}\) -\(\frac{4}{3}\)
=\(\frac{3}{4}\) + \(\frac{14}{3}\) -\(\frac{4}{3}\)
=\(\frac{9}{12}\) +\(\frac{56}{12}\)
=\(\frac{56}{12}\) - \(\frac{4}{3}\)
=\(\frac{56}{12}\)- \(\frac{16}{12}\)
=\(\frac{10}{3}\)
b) \(\frac{5}{2}\) - [\(\frac{1}{2}\) + (\(\frac{-2}{3}\)) .\(\frac{6}{5}\)].\(\frac{2}{5}\)
= \(\frac{5}{2}\) - [\(\frac{1}{2}\) + (\(\frac{-4}{5}\))].\(\frac{2}{5}\)
= \(\frac{5}{2}\) - (\(\frac{5}{10}\) +\(\frac{-8}{10}\)).\(\frac{2}{5}\)
=\(\frac{5}{2}\)-\(\frac{3}{10}\).\(\frac{2}{5}\)
=\(\frac{5}{2}\)-\(\frac{3}{25}\)
= \(\frac{125}{50}\)-\(\frac{6}{50}\)
= \(\frac{119}{50}\)
(Mình không chắc chắn lắm!^^)
Bài làm:
Ta có: \(\left(\frac{1}{2}x+\frac{3}{4}\right)^2\div\left(-\frac{3}{14}\right)=-\frac{7}{24}\)
\(\Leftrightarrow\left(\frac{1}{2}x+\frac{3}{4}\right)^2=\frac{1}{16}=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x+\frac{3}{4}=\frac{1}{4}\\\frac{1}{2}x+\frac{3}{4}=-\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=-\frac{1}{2}\\\frac{1}{2}x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
(\(\frac{1}{2}\).x+\(\frac{3}{4}\))2: \(\frac{-3}{14}\)=\(\frac{-7}{24}\)
(\(\frac{1}{2}\).x+\(\frac{3}{4}\))2 =\(\frac{1}{16}\)
(\(\frac{1}{2}\).x+\(\frac{3}{4}\))2=(\(\frac{1}{4}\))2
\(\hept{\begin{cases}\frac{1}{2}.x+\frac{3}{4}=\frac{1}{4}\\\frac{1}{2}.x+\frac{3}{4}=\frac{-1}{4}\end{cases}}\)<=>\(\hept{\begin{cases}\frac{1}{2}.x=\frac{1}{4}-\frac{3}{4}\\\frac{1}{2}.x=\frac{-1}{4}-\frac{3}{4}\end{cases}}\)<=>\(\hept{\begin{cases}\frac{1}{2}.x=-\frac{1}{2}\\\frac{1}{2}.x=-1\end{cases}}\)
<=>\(\hept{\begin{cases}x=\frac{-1}{2}:\frac{1}{2}=-1\\x=-1:\frac{1}{2}=-2\end{cases}}\)
Lời giải:
$7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
Bạn ơi link nè, vô đây xem chắc chắn sẽ có thể đổi tên nha bạn: https://www.youtube.com/watch?v=cPQQs5SmzgY