Question

[Trần Quang Minh]

1/2(x+1)(3-x)+x=3

Mng giúp mình với ạ

Answer 1

[Toru]

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy phương trình đã cho có tập nghiệm là \(S=\left\{3;1\right\}\).

$Toru$

Answer 2

[Nguyễn Lê Phước Thịnh]

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

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

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

=>\(-x^2+2x+3+2x=6\)

=>\(-x^2+4x-3=0\)

=>\(\left(x-1\right)\left(x-3\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)