#)Giải :
a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(5x=\frac{6}{7}-1\)
\(5x=\frac{6}{7}-\frac{7}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{7}\div5\)
\(x=-\frac{1}{7}\times\frac{1}{5}\)
\(x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
a) (5x + 1)2 = 36/49
<=> (5x + 1)2 = (6/7)2
<=> \(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)
<=>\(\orbr{\begin{cases}5x=\frac{6}{7}-1\\5x=-\frac{6}{7}-1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{cases}}\)
Vậy: ...
\(a,\left(5x-1\right)^2=\frac{36}{49}\)
\(\Rightarrow\left(5x+1\right)^2\)
\(=\left(\frac{6}{7}\right)^2\Rightarrow5x+1\)
\(=\frac{6}{7}\Rightarrow5x\)
\(=-\frac{1}{7}\Rightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{3}\right)^3=\left(\frac{2^2}{3^2}\right)^3\)
\(\Leftrightarrow\left(x-\frac{2}{3}\right)=\frac{2^2}{3^2}\)
\(x-\frac{2}{3}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{2}{3}\)
\(x=\frac{4}{9}+\frac{6}{9}\)
\(x=\frac{10}{9}\)
Vậy \(x=\frac{10}{9}\)
b) (x - 2/9)3 = (2/3)6
<=> (x - 2/9)3 = 23/33
<=> (x - 2/9)3 = 8/27
<=> (x - 2/9)3 = (2/3)3
<=> x - 2/9 = 2/3
<=> x = 2/3 + 2/9
<=> x = 8/9
=> x = 8/9
\(a,\text{ }\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\pm\frac{6}{7}\right)^2\)
\(5x+1=\pm\frac{6}{7}\)
\(\Rightarrow\orbr{\begin{cases}5x+1=-\frac{6}{7}\\5x+1=\frac{6}{7}\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}5x=-\frac{6}{7}-1=-\frac{13}{7}\\5x=\frac{6}{7}-1=-\frac{1}{7}\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{7}\text{ : }5=-\frac{13}{35}\\x=-\frac{1}{7}\text{ : }5=-\frac{1}{35}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-\frac{13}{35}\text{ ; }-\frac{1}{35}\right\}\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)
\(\left(x-\frac{2}{9}\right)=\left(\frac{4}{9}\right)^3\)
\(x-\frac{2}{9}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{2}{9}\)
\(x=\frac{2}{3}\)
\(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(8x-1=5\)
\(8x=5+1\)
\(8x=6\)
\(x=\frac{3}{4}\)
