(5x+1)^2 = 36/49

5x+ 1 = 6/7

5x = -1/7

x = -1/35

(5x+1)²=36/49

Mà: (5x+1)²=36/49=(6/7)² hoặc (5x+1)²=36/49=(-6/7)²

=>5x+1=6/7 hoặc 5x+1= -6/7

=> 5x= 6/7-1 hoặc 5x= -6/7 -1

=> 5x= -1/7 hoặc 5x= -13/7

=> x= -1/7: 5 hoặc x= -13/7: 5

=> x=-1/35 hoặc x= -13/35

(5x+1)²=36/49

Mà: (5x+1)²=36/49= (6/7)²

=>5x+1=6/7

=>5x=6/7-1

=>5x=-1/7

=>x= -1/7 : 5

=>x= -1/ 35

ta có:

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

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

\(\Rightarrow5x+1=\frac{6}{7}\Rightarrow5x=\frac{-1}{7}\Rightarrow x=\frac{-1}{35}\)

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

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

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

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

\(x=\frac{-1}{3}:5=\frac{-1}{3}.\frac{1}{5}=\frac{-1}{15}\)

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

(+) TH 1: 5x + 1 = 6/7

              5x        = 6/7 - 1

               5x        =  -1/7

                 x        = -1/7 : 5 

                 x          = -1 /35

(+) TH2 : 5x + 1 = - 6/7

             5x         = -6/7 - 1

             5x         = -13/7

               x         =-13/7 : 5 

               x          = -13/35

ko ghi lại đề bài lm luôn:

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

       \(5x+1=\frac{6}{7}\)

                \(5x=\frac{-1}{7}\)

                  \(x=\frac{-1}{35}\)

tôi lm đại thôi nhưng chắc đ đấy

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=\(\dfrac{-13}{35}\)

( 5x + 1 )² = 36/49

<=> ( 5x + 1 )² = ( ±6/7 )²

<=> 5x + 1 = 6/7 hoặc 5x + 1 = -6/7

<=> x = -1/35 hoặc x = -13/35

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

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

\(\Rightarrow\orbr{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}\)

\(\left(5x+1\right)=\pm\sqrt{\frac{36}{49}}=\pm\frac{6}{7}\)                         

\(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)      

\(\orbr{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}}\)    

\(\orbr{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}\)

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

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

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

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

\(\Rightarrow x=-\frac{1}{7}:5\)               \(\Rightarrow x=-\frac{1}{7}\cdot\frac{1}{5}=-\frac{1}{35}\)

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

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

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

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

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

Vậy ...

 Ta có: \(\left(5x+1\right)^2=\frac{36}{49}=\left(\pm\frac{6}{7}\right)^2\)

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

 ​TH1: \(5x+1=\frac{6}{7}\Rightarrow5x=\frac{6}{7}+1=\frac{13}{7}\Rightarrow x=\frac{13}{7}:5=\frac{13}{35}\)

TH2: \(5x+1=\frac{-6}{7}\Rightarrow5x=\frac{-6}{7}+1=\frac{1}{7}\Rightarrow x=\frac{1}{7}:5=\frac{1}{35}\)

 Vậy: \(x\in\left\{\frac{13}{35};\frac{1}{35}\right\}\)