#)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: ...

1.

a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)

b) x=0

d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)

e) \(x=\frac{2}{3}\)

a: =>5x+1=6/7 hoặc 5x+1=-6/7

=>5x=-1/7 hoặc 5x=-13/7

=>x=-1/35 hoặc x=-13/35

b: =>x-2/9=4/9

=>x=6/9=2/3

c: =>8x+1=5

=>8x=4

hay x=1/2

a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{9}\right)\)\(^2\)

\(5x+1=\pm\dfrac{6}{9}\)

+) \(5x+1=\dfrac{6}{9}\)

\(5x=\dfrac{6}{9}-1=\dfrac{6}{9}-\dfrac{9}{9}\)

\(5x=\dfrac{-5}{9}\)

\(x=\dfrac{-5}{9}:5=\dfrac{-1}{45}\)

+) \(5x+1=\dfrac{-6}{9}\)

\(5x=\dfrac{-6}{9}-1=\dfrac{-6}{9}-\dfrac{9}{9}\)

\(5x=\dfrac{-5}{3}\)

\(x=\dfrac{-5}{3}:5=\dfrac{-5}{15}\)

vậy \(x\in\left\{\dfrac{-5}{15};\dfrac{-1}{45}\right\}\)

a/ ĐKXĐ: ...

\(\frac{5x-3}{2\left(5x-1\right)\left(5x+1\right)}-\frac{5x-9}{12x\left(5x-1\right)}+\frac{1}{12x}=\frac{8x-5}{16x\left(5x+1\right)}\)

\(\Leftrightarrow\frac{5x-3}{\left(5x-1\right)\left(5x+1\right)}+\frac{1}{6x}\left(1-\frac{5x-9}{5x-1}\right)=\frac{8x-5}{8x\left(5x+1\right)}\)

\(\Leftrightarrow\frac{5x-3}{\left(5x-1\right)\left(5x+1\right)}+\frac{4}{3x\left(5x-1\right)}-\frac{8x-5}{8x\left(5x+1\right)}=0\)

\(\Leftrightarrow24x\left(5x-3\right)+32\left(5x+1\right)-3\left(5x-1\right)\left(8x-5\right)=0\)

\(\Leftrightarrow-120x^2+379x-55=0\)

Bạn có nhầm đề chỗ nào ko nhỉ? Con số thật khủng khiếp (nghiệm ko hề đẹp)

b/ ĐKXĐ:...

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\right)=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+9}=27-\frac{1}{x+9}\)

\(\Leftrightarrow\frac{1}{x}=27\Rightarrow x=\frac{1}{27}\)

a, \(\frac{5x-3}{50x^2-2}+\frac{5x-9}{12x-60x^2}+\frac{1}{12x}=\frac{8x-5}{80x^2+16x}\) (ĐKXĐ: x \(\ne\) \(\pm\)\(\frac{1}{5}\); x \(\ne\) 0)

\(\Leftrightarrow\) \(\frac{5x-3}{2\left(5x-1\right)\left(5x+1\right)}+\frac{-5x+9}{12x\left(5x-1\right)}+\frac{1}{12x}=\frac{8x-5}{16x\left(5x+1\right)}\)

\(\Leftrightarrow\) \(\frac{24x\left(5x-3\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}+\frac{-4\left(5x+1\right)\left(5x-9\right)}{48x\left(5-1x\right)\left(5x+1\right)}+\frac{4\left(5x-1\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}=\frac{3\left(8x-5\right)\left(5x-1\right)}{48x\left(5x-1\right)\left(5x+1\right)}\)

\(\Leftrightarrow\) 24x(5x - 3) - 4(5x + 1)(5x - 9) + 4(5x - 1)(5x + 1) = 3(8x - 5)(5x - 1)

\(\Leftrightarrow\) 120x² - 72x - 100x² + 160x + 36 + 100x² - 4 = 120x² - 99x + 15

\(\Leftrightarrow\) 120x² - 120x² - 100x² + 100x² - 72x + 160x + 99x = 15 - 36 + 4

\(\Leftrightarrow\) 187x = -17

\(\Leftrightarrow\) x = \(\frac{-1}{11}\) (TM ĐKXĐ)

Vậy S = {\(\frac{-1}{11}\)}

Chúc bn học tốt!! (Đã được kiểm chứng không sai :)

b, \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) -3; x \(\ne\) -6; x \(\ne\) -9)

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) - \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}-27+\frac{1}{x+9}\)) = 0

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-27\)) = 0

\(\Leftrightarrow\) \(\frac{1}{x}-27\) = 0

\(\Leftrightarrow\) x = \(\frac{1}{27}\) (TM ĐKXĐ)

Vậy S = {\(\frac{1}{27}\)}

Chúc bn học tốt!!

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).