Question

[Hà Thúy Quỳnh]

a, (x+2018)(1/2+2/7)=(x+2018).(1/5+1/6)

b, 7(x-1)+2x(x-1)=0

Answer 1

[Lê Thanh Nhàn]

a) \(\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}\right)=\left(x+2018\right)\left(\frac{1}{5}+\frac{1}{6}\right)\)

\(\Leftrightarrow\) \(\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}\right)-\left(x+2018\right)\left(\frac{1}{5}+\frac{1}{6}\right)\) = 0

\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow x+2018=0\)

\(\Leftrightarrow x=-2018\)

b) \(7\left(x-1\right)+2x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7+2x\right)=0\)

\(\Leftrightarrow\) x - 1 = 0 hoặc 7 + 2x = 0

1) x - 1 = 0 \(\Leftrightarrow\) x = 1

2) 7 + 2x = 0 \(\Leftrightarrow\) -3,5

Vậy: x = 1; -3,5

Answer 2

[Vũ Minh Tuấn]

b) \(7\left(x-1\right)+2x\left(x-1\right)=0\)

=> \(\left(x-1\right).\left(7+2x\right)=0\)

=> \(\left\{{}\begin{matrix}x-1=0\\7+2x=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=0+1\\2x=0-7=-7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1\\x=\left(-7\right):2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=1\\x=-\frac{7}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{1;-\frac{7}{2}\right\}.\)

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