Question

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

Answer 1

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

Answer 2

(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

Answer 3

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

=>\(\left(5x+1\right)^2=\left(\dfrac{6}{7}\right)^2\)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=\(\dfrac{-13}{35}\)

Answer 4

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

=> \(\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\left(\dfrac{6}{7}\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}5x=-\left(\dfrac{1}{7}\right)\\5x=-\left(\dfrac{13}{7}\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\left(\dfrac{1}{35}\right)\\x=-\left(\dfrac{13}{35}\right)\end{matrix}\right.\)

Answer 5

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

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

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

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

\(\Leftrightarrow x=\dfrac{-1}{35}\)

Answer 6

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

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

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

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

\(\left[{}\begin{matrix}x=\dfrac{13}{35}\\x=\dfrac{1}{35}\\x=\dfrac{-13}{35}\\x=\dfrac{-1}{35}\end{matrix}\right.\)