Toán Lớp 8: giải phương trình
$\frac{4+x}{3}+$ $\frac{\frac{x-1}{2}-x}{2}+\frac{x-3}{4}=\frac{6-2x}{2}$
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Toán Lớp 8: giải phương trình
$\frac{4+x}{3}+$ $\frac{\frac{x-1}{2}-x}{2}+\frac{x-3}{4}=\frac{6-2x}{2}$
Comments ( 1 )
$\dfrac{4+x}{3}+\dfrac{\dfrac{x-1}{2}-x}{2}+\dfrac{x-3}{4}=\dfrac{6-2x}{2}$
$⇔\dfrac{4+x}{3}\cdot \:12+\dfrac{\dfrac{x-1}{2}-x}{2}\cdot \:12+\dfrac{x-3}{4}\cdot \:12=\dfrac{6-2x}{2}\cdot \:12$
$⇔4\left(x+4\right)+3\left(-x-1\right)+3\left(x-3\right)=12\left(3-x\right)$
$⇔4x+4=36-12x$
$⇔4x+4-4=36-12x-4$
$⇔4x=-12x+32$
$⇔4x+12x=-12x+32+12x$
$⇔16x=32$
$⇔x=\dfrac{32}{16}$
$⇔x=2$
$\text{Vậy pt có nghiệm S = {2}}$