Question

[Tùng Hoàng]

(x^2+x+1).3x+1/x+2=(x^2+x+1).x/2(x+2) Giup minh voi

Answer 1

[YangSu]

\(\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}=\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}\) \(\left(dkxd:x\ne-2\right)\)

\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)

Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm

\(\Leftrightarrow6x+2-x=0\)

\(\Leftrightarrow5x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)

Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)

Answer 2

[Hobiee]

\(\dfrac{\left(x^2+x+1\right).\left(3x+1\right)}{x+2}=\dfrac{\left(x^2+x+1\right).x}{2\left(x+2\right)}\)

hay \(\left(x^2+x+1\right).\dfrac{3x+1}{x+2}=\left(x^2+x+1\right).\dfrac{x}{2\left(x+2\right)}\)