\(\frac{5x+3}{1}=\frac{\frac{7}{15}}{5x+3}\)
Ta có:
\(\left(5x+3\right).\left(5x+3\right)=\frac{7}{15}.1\)
\(2.\left(5x+3\right)=\frac{7}{15}\)
\(5x+3=\frac{7}{15}:2\)
\(5x+3=\frac{7}{30}\)
\(5x=\frac{7}{30}-3\)
\(5x=\frac{-83}{30}\)
\(x=-\frac{83}{30}:5=\frac{-83}{150}\)
Vậy \(x=\frac{-83}{150}\)
Công: Em bị sai từ dòng 3 rồi ^^ Xem lại nhé :)
Cô giải như sau: Đk: \(x\ne-\frac{3}{5}\)
\(\frac{5x+3}{1}=\frac{\frac{7}{15}}{5x+3}\Leftrightarrow\left(5x+3\right)\left(5x+3\right)=\frac{7}{15}\Leftrightarrow\left(5x+3\right)^2=\frac{7}{15}\)
Do \(\left(\sqrt{\frac{7}{15}}\right)^2=\frac{7}{15}\) và \(\left(-\sqrt{\frac{7}{15}}\right)^2=\frac{7}{15}\) nên ta có hai trường hợp:
TH1: \(5x+3=\sqrt{\frac{7}{15}}\Rightarrow x=\frac{\sqrt{\frac{7}{15}}-3}{5}\left(tmđk\right)\)
TH2: \(5x+3=-\sqrt{\frac{7}{15}}\Rightarrow x=\frac{-\sqrt{\frac{7}{15}}-3}{5}\left(tmđk\right)\)
Công: Sai ở chỗ em nhầm \(\left(5x+3\right)\left(5x+3\right)với\left(5x+3\right)+\left(5x+3\right)\)
Chỉ \(\left(5x+3\right)+\left(5x+3\right)=2\left(5x+3\right)\) , còn \(\left(5x+3\right)\left(5x+3\right)=\left(5x+3\right)^2\) nhé em ^^
