Question

[Đỗ Huệ Anh]

Mình sẽ tick cho tất cả những ai trả lời mình ! Giúp mình nha.

Answer 1

[Athanasia Karrywang]

\(a,x+6x^2=0\)

=>  \(x\left(1+6x\right)=0\)

=>  \(\orbr{\begin{cases}x=0\\1+6x=0\Leftrightarrow6x=-1\Leftrightarrow x=-\frac{1}{6}\end{cases}}\)

\(b,2\left(x+3\right)-x\left(x+3\right)=0\)

=>  \(\left(2-x\right)\left(x+3\right)=0\)

=>  \(\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

\(c,5x\left(x-2\right)-\left(2-x\right)=0\)

=>  \(\left(5x+2\right)\left(x-2\right)=0\)

=>  \(\orbr{\begin{cases}5x+2=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{5}\\x=2\end{cases}}}\)

\(d,\left(x+1\right)=\left(x+1\right)^2\)

=> \(\left(x+1\right)^2-\left(x+1\right)=0\)

=>   \(\left(x+1\right)\left(x+1-1\right)=0\)

=>  \(\orbr{\begin{cases}x+1=0\\x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)