Question

BC/x = BC+6x/BC => BC=3x

Mọi người giải thích các bước làm hộ em với ạ^^ Em cảm ơn!! 

Answer 1

\(\dfrac{BC}{x}=\dfrac{BC+6x}{BC}=>BC^2=BC.x+6x^2\)

\(=>6x^2+BC.x-BC^2=0\)

\(< =>6\left(x^2+\dfrac{1}{6}BCx-\dfrac{1}{6}BC^2\right)=0\)

\(=>x^2+\dfrac{1}{6}BCx-\dfrac{1}{6}BC^2=0\)

\(< =>x^2+2.\dfrac{1}{12}BC.x+\left(\dfrac{1}{12}BC^2\right)-\left(\dfrac{1}{12}BC\right)^2-\dfrac{1}{6}BC^2=0\)

\(< =>\left(x+\dfrac{1}{12}BC\right)^2-\left(\dfrac{5}{12}BC\right)^2=0\)

\(=>\left(x+\dfrac{1}{12}BC+\dfrac{5}{12}BC\right)\left(x+\dfrac{1}{12}BC-\dfrac{5}{12}BC\right)=0\)

\(< =>\left(x+\dfrac{1}{2}BC\right)\left(x-\dfrac{1}{3}BC\right)=0\)

\(=>\left[{}\begin{matrix}x+\dfrac{1}{2}BC=0\\x-\dfrac{1}{3}BC=0\end{matrix}\right.=>\left[{}\begin{matrix}BC=2x\\BC=3x\end{matrix}\right.\)