Bài 1: Tìm x,biết
a,x- $\frac{-5}{3}$ = $\frac{-2}{8}$
b, $\frac{x-3}{-2}$ =$\frac{-32}{x-3}$
Bài 1: Tìm x,biết a,x- $\frac{-5}{3}$ = $\frac{-2}{8}$ b, $\frac{x-3}{-2}$ =$\frac{-32}{x-3}$
By Autumn
By Autumn
Bài 1: Tìm x,biết
a,x- $\frac{-5}{3}$ = $\frac{-2}{8}$
b, $\frac{x-3}{-2}$ =$\frac{-32}{x-3}$
Bài làm :
a,
`x-(-5)/3=(-2)/8`
`->x+5/3=(-1)/4`
`->x=(-1)/4-5/3`
`->x=(-3)/12-20/12`
`->x=-23/12`
Vậy `x=-23/12`
b,
`(x-3)/(-2)=(-32)/(x-3)` Điều kiện : `x \ne 3`
`->(x-3)^2=(-2).(-32)`
`->(x-3)^2=64`
`->(x-3)^2=(\pm 8)^2`
`->` \(\left[ \begin{array}{l}x-3=8\\x-3=-8\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=11\\x=-5\end{array} \right.\)
Vậy `x \in {11;-5}`
`Giải`
$a)x-$ $\dfrac{-5}{3}$ $=$ $\dfrac{-2}{8}$
$x=$ $\dfrac{-2}{8}$ $+$ $\dfrac{-5}{3}$
$x=$ $\dfrac{-23}{12}$
$Vậy$ $x=$ $\dfrac{-23}{12}$
$b)$ $\dfrac{x-3}{-2}$ $=$ $\dfrac{-32}{x-3}$
$=> (x-3).(x-3)$ $=(-2).(-32)$
$=>$ $(x-3)^{2}$ $=(-2).(-32)$
$=>$ $(x-3)^{2}$ $=64$
$=>$ $(x-3)^{2}$ $=$$8^{2}$ $hoặc =$ $(-8)^{2}$ $(x$$\neq$ $0)$
$Ta$ $có$ $hai$ $trường$ $hợp :$
$Th1:$ $x-3=8$
$x=11$ $( thỏa$ $mãn)$
$Th2:$ $x-3=-8$
$x=-5$ $( thỏa$ $mãn)$
$Vậy$ $x=11$ $hoặc$ $x=-5$