Các bước giải chi tiết
a) \(\dfrac{392-x}{32}\) + \(\dfrac{390-x}{34}\) + \(\dfrac{388-x}{36}\) = -3
⇔ \(\dfrac{392-x}{32}\)+1+\(\dfrac{390-x}{34}\)+1+\(\dfrac{388-x}{36}\)+1 = 0
⇔\(\dfrac{424-x}{32}\)+\(\dfrac{424-x}{34}\)+\(\dfrac{424-x}{36}\)=0
⇔\(\left(424-x\right)\)\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\right)\)=0
⇔\(424-x\) = 0\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\ne\forall x\right)\)
⇔\(x=424\)
b) \(\dfrac{x-3}{3}\)- \(x\) = \(5-\dfrac{x+1}{4}\)
⇔\(\dfrac{x-3-3x}{3}\) = 5 + \(\dfrac{-\left(x+1\right)}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = 5 + \(\dfrac{-x-1}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{20-x-1}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{-x+19}{4}\)
⇔ \(4\left(-2x-3\right)\) = \(3\left(-x+19\right)\)
⇔\(-8x-12\) = \(-3x+57\)
⇔\(-8x\) = \(-3x+69\)
⇔\(-5x=69\)
⇔ \(x=-\dfrac{69}{5}\)
(392-x)/32+(390-x)/34+(388-x)/36=-3
=>392-x)/32 +1 + (390-x)/34 +1 +(388-x)/36 +1=0
=>(424-x)/32+(424-x)/34+(424-x)/36=0
=>424-x=0(vì 1/32+1/34+1/36 khác 0)
=>x=424