Question

[Kochou Shinobu]

Tìm\(x\inℚ\)thỏa mãn:

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)

Giúp mk vs

Cảm ơn nhiều ạ!!

Answer 1

[๖²⁴ʱ๖ۣۜTɦủү❄吻༉]

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)

\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)

Khử mẫu : \(12x-12+6x-6=4x+3x-7\)

\(\Leftrightarrow18x-18=7x-7\Leftrightarrow11x=11\Leftrightarrow x=1\)

Answer 2

[Khánh Ngọc]

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)

\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)

\(\Leftrightarrow\frac{12x-12+6x-6}{12}=\frac{4x+3x-7}{12}\)

\(\Leftrightarrow18x-18=7x-7\)

\(\Leftrightarrow18x+7x=18+7\)

\(\Leftrightarrow25x=25\)

\(\Leftrightarrow x=1\)