Question

[Đặng Viết Hoàng]

tìm x biết : x+1/2=8/x+1

Answer 1

[subjects]

\(\dfrac{x+1}{2}=\dfrac{8}{x+1}\\ =>\left(x+1\right)^2=2\cdot8\\ =>\left(x+1\right)^2=16=4^2=\left(-4\right)^2\\ x+1=4=>x=3\\ x+1=-4=>x=-5\)

vậy x ∈ {3; -5}

Answer 2

[Bùi Ngọc Hà]

x+1/2​=8/x+1
x+1/2-1=8/x
x−1/2=8/x
x⋅(x−1/2​)=x⋅8/x
x2−x/2​=8
2x2−x=16
2x2−x−16=0
x=2(2)−(−1)±(−1)2−4(2)(−16)​​ \(x = \frac{1 \pm \sqrt{1 + 128}}{4}\) \(x = \frac{1 \pm \sqrt{129}}{4}\)
x=41+129​​hoặcx=41−129​​