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

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

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

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

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

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

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

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

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

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

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

Mấy bạn làm sai rồi, mũ chẵn chia 2 trường hợp, mũ lẻ chia 1 trường hợp nhé.

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

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

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

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

Mk quên mất chia 2 TH

Sorry!!!

5x=8x và y-x =-12

x/y =7/13 và y+x = 40

Ken Kun

5x = 8x???!!!!!

+) \(\frac{x}{y}=\frac{7}{13}\)và y + x = 40

Ta có: \(\frac{x}{y}=\frac{7}{13}\Rightarrow\frac{x}{7}=\frac{y}{13}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{7}=\frac{y}{13}=\frac{x+y}{7+13}=\frac{40}{20}=2\)

\(\Rightarrow\frac{x}{7}=2\Rightarrow x=14\)

\(\frac{y}{13}=2\Rightarrow y=26\)

Vậy x = 14; y = 26