Question

Bài 1 : tìm x 
\(a,\left(5x-3\right)^2-\frac{1^2}{64}=0\)
\(b,3x-7\left(5x-1\right)=6-2\left(4-3x\right)\)
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Answer 1

Ta có : \(\left(5x-3\right)^2-\frac{1^2}{64}=0\)

\(\Leftrightarrow\left(5x-3\right)^2=\frac{1}{64}\)

\(\Leftrightarrow\left(5x-3\right)^2=\left(\frac{1}{8}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=-\frac{1}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{1}{8}+3\\5x=-\frac{1}{8}+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25}{8}.\frac{1}{5}\\x=\frac{23}{8}.\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)

Answer 2

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

Answer 3

\(\frac{1}{5}\)ở đâu ra vậy bạn

Answer 4

\(\left(5x-3\right)^2-\frac{1}{64}=0\)

\(\left(5x-3\right)^2=\left(\frac{1}{16}\right)^2\)

\(5x=\frac{1}{16}+3=\frac{49}{16}\)

\(x=\frac{49}{80}\)