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

\(\Rightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2=\left(\frac{-6}{7}\right)^2\)

\(\Rightarrow\hept{\begin{cases}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{cases}}\Rightarrow\hept{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}}\)

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}^2\right)^3\)

\(\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)

\(x-\frac{2}{9}=\frac{4}{9}\)

\(x=\frac{4}{9}+\frac{2}{9}\)

\(x=\frac{6}{9}=\frac{2}{3}\)

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

\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)

\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

\(\Rightarrow5x+1=\frac{6}{7}\)

\(\Rightarrow5x=\frac{6}{7}-1\)

\(\Rightarrow5x=-\frac{1}{7}\)

\(\Rightarrow x=-\frac{1}{7}:5\)

\(\Rightarrow x=-\frac{1}{35}\)

Vậy \(x=-\frac{1}{35}\)

\(b\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right).^6\)

\(\left(x-\frac{2}{9}\right)^3=\frac{\left(2^2\right)^3}{\left(3^2\right)^3}\)

\(\left(x-\frac{2}{9}\right)^3=\frac{4^3}{9^3}\)

\(\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)

\(\Rightarrow x-\frac{2}{9}=\frac{4}{9}\)

\(\Rightarrow x=\frac{4}{9}+\frac{2}{9}\)

\(\Rightarrow x=\frac{6}{9}=\frac{2}{3}\)

Vậy \(x=\frac{2}{3}\)

a) (5x + 1)² = 36/49

     (5x + 1)² = (6/7)²

suy ra: 5x + 1= 6/7               hoặc         5x +1 = -6/7

                 5x    = 6/7 - 1                               5x  = -6/7 -1

                 5x   = -1/7                                    5x   = -13/7

                   x    =  -1/7 : 5                               x   = -13/7 : 5

                   x   = -1/35                                   x     = -13/35

a,   \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
=> 5x + 1 = \(\frac{6}{7}\)
\(\Rightarrow5x=\frac{6}{7}-1\)
\(\Rightarrow5x=\frac{-1}{7}\)
\(\Rightarrow x=\frac{-1}{7}:5=\frac{-1}{35}\)
b,,   \(\left(2x-\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\left(2x-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2\)
\(\Rightarrow2x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow2x=\frac{1}{2}+\frac{1}{2}=1\)
\(\Rightarrow x=1:2=\frac{1}{2}\)

b) (2x - 1/2 )² = (1/2)²

   2x - 1/2  = 1/2                     hoặc              2x - 1/2 = -1/2

           2x = 1/ 2  + 1/2                                        2x  = -1/2 + 1/2

            2x  = 1                                                      2x  = 0

            x = 1/2                                                      x     = 0   

#)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: ...

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

\(\Leftrightarrow\left(5x+1\right)^2=\left(\dfrac{6}{7}\right)^2\)

\(\Leftrightarrow5x+1=\dfrac{6}{7}\)

\(\Leftrightarrow5x=\dfrac{13}{7}\)

\(\Leftrightarrow x=\dfrac{13}{35}\)

Vậy \(x=\dfrac{13}{35}\)

(5x + 1)² = \(\dfrac{36}{49}\)
<=> (5x + 1)² = (\(\dfrac{6}{7}\))²
<=> 5x + 1 = \(\dfrac{6}{7}\)
<=> 5x = \(-\dfrac{1}{7}\)
<=> x = \(-\dfrac{1}{35}\)
@Võ Ngọc Tường Vy

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

=>\(\left(5x+1\right)^2=\left(\dfrac{6}{7}\right)^2\)hoặc \(\left(5x+1\right)^2=\left(\dfrac{-6}{7}\right)^2\)

=>5x+1=\(\dfrac{6}{7}\)hoặc 5x+1=\(\dfrac{-6}{7}\)

=>5x=\(\dfrac{-1}{7}\)hoặc 5x=\(\dfrac{-13}{7}\)

=>x=\(\dfrac{-1}{35}\)hoặc x=\(\dfrac{-13}{35}\)

a)\(5^x.\left(5^3\right)^2=625\)

\(5^x.5^6=5^4\)

\(5^x=5^{-2}\)

\(x=-2\)

b)\(27< 81^3:3^x< 243\)

\(3^3< \left(3^4\right)^3:3^x< 3^5\)

\(3^3< 3^{12}:3^x< 3^5\)

\(3^{12}:3^x=3^4\)

\(3^x=3^3\)

\(x=3\)

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

\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

\(5x+1=\frac{6}{7}\)

\(5x=\frac{-1}{7}\)

\(x=\frac{-1}{35}\)

d)\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)

\(x-\frac{2}{9}=\frac{4}{9}\)

\(x=\frac{6}{9}=\frac{2}{3}\)

\(5^x.\left(5^3\right)^2=625\)

\(\Rightarrow5^x.5^6=5^4\)

\(\Rightarrow5^{x+6}=5^4\Rightarrow x+6=4\Rightarrow x=-2\)

Đề sai rồi bạn : Phải là :

 \(5^x:\left(5^3\right)^2=625\)

\(\Rightarrow5^x:5^6=5^4\)

\(\Rightarrow5^{x-6}=5^4\)

\(\Rightarrow x-6=4\Rightarrow x=10\)

Nhứng nếu đề đúng thì bạn có thể lấy KQ trên

a) (5x +1)^2= 6^2/7^2

=> 5x+1= 6/7 hoặc -6/7 ( vì cả hai đều có mũ hai nên có thể bỏ đi - cái này mình giải thích cho bạn hỉu thui, đừng chép vào vở nhé)

 Đến đây thì bạn cứ tính theo cách tìm x thông thường, cuối cùng thì ra số âm nên không có kết quả x thuộc N

Tiếc quá chưa học đến

sory bạn

cô lên

a) (5x +1 ) ² = 36²/49²

=> (5x + 1 )   = 36/49

=> 5x            = 36/49 - 1 = -13/49

=> x              = -13/245