Question

Tìm x: \(5x^2+7,1=\sqrt{49}\)

Answer 1

\(5x^2+7,1=\text{√}49\)

\(\Rightarrow5x^2+7,1=7\)

\(\Rightarrow5x^2=7-7,1=-0,1\)

\(\Rightarrow x^2=\left(-0,1\right):5=\left(-0,02\right)\)

\(\Rightarrow x\in\varnothing\)

Answer 2

\(5x^2+7,1=\sqrt{49}\)

\(\Rightarrow5x^2+7,1=7\)

\(\Rightarrow5x^2=-0,1\)

\(\Rightarrow x^2=-0,1:5\Rightarrow x^2=-0,02\Rightarrow x=-\sqrt{0,02}\) hoặc \(x=\sqrt{0,02}\)

Vậy x=\(\sqrt{0,02}\)hoặc \(x=-\sqrt{0,02}\)