giải phương trình sau x+4/8+x+6/7=x-2/11+x-4/12
2 câu trả lời
#andy
\[\begin{array}{l}
\dfrac{{x + 4}}{8} + \dfrac{{x + 6}}{7} = \dfrac{{x - 2}}{{11}} + \dfrac{{x - 4}}{{12}}\\
MC:1848\\
\Leftrightarrow \dfrac{{231\left( {x + 4} \right)}}{{8.231}} + \dfrac{{264\left( {x + 6} \right)}}{{7.264}} = \dfrac{{168\left( {x - 2} \right)}}{{11.168}} + \dfrac{{154\left( {x - 4} \right)}}{{154.12}}\\
\Leftrightarrow \dfrac{{231x + 924}}{{1848}} + \dfrac{{264x + 1584}}{{1848}} = \dfrac{{168x - 336}}{{1848}} + \dfrac{{154x - 616}}{{1848}}\\
\Leftrightarrow 231x + 924 + 264x + 1584 = 168x - 336 + 154x - 616\\
\Leftrightarrow 231x + 264x - 168x - 154x = - 336 - 616 - 924 - 1584\\
\Leftrightarrow 173x = - 3460\\
\Leftrightarrow x = - 20
\end{array}\]
`(x+4)/8+(x+6)/7=(x-2)/11+(x-4)/12`
`<=>x/8+1/2+x/7+6/7=x/11-2/11+x/12-1/3`
`<=>x(1/8 +1/7) + 19/14 = x(1/11 + 1/12)-17/33`
`<=>15/56x+19/14=23/132x-17/33`
`<=>23/132x-15/56x=17/33+19/14`
`<=>(-173)/1848x=865/462`
`<=>x=-20`
Vậy phương trình có tập nghiệm `S={-20}`