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Ta có : \(\left(5x-3\right)^2-\frac{1^2}{64}=0\)
\(\Leftrightarrow\left(5x-3\right)^2=\frac{1}{64}\)
\(\Leftrightarrow\left(5x-3\right)^2=\left(\frac{1}{8}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=-\frac{1}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{1}{8}+3\\5x=-\frac{1}{8}+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25}{8}.\frac{1}{5}\\x=\frac{23}{8}.\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)
b) 3x - 7.(5x-1) = 6 - 2.(4-3x)
=> 3x - 35x + 7 = 6 - 8 + 6x
=> 3x - 35x - 6x = 6-8 -7
-38x = -9
x = 9/38
A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{30}{31}\)
=\(\frac{-671}{4805}\)
a) \(\frac{3}{7}x-\frac{1}{35}=\frac{3}{5}\)
\(\frac{3}{7}x=\frac{3}{5}+\frac{1}{35}\)
\(\frac{3}{7}x=\frac{22}{35}\)
\(x=\frac{49}{35}=1,4\)
b) \(1,5-x:\frac{1}{2}=\frac{1}{4}\)
\(x:\frac{1}{2}=1,5-\frac{1}{4}\)
\(x:\frac{1}{2}=\frac{5}{4}\)
\(x=\frac{5}{4}.\frac{1}{2}\)
\(x=\frac{5}{8}\)
Vậy ..