Question

[lư thị ngọc giao]

\(\sqrt{x+3}+\sqrt{4x+12}=5\)  giai phuong trinh

Answer 1

[Thắng Nguyễn]

\(\sqrt{x+3}+\sqrt{4x+12}=5\)

Đk:\(x\ge3\)

\(pt\Leftrightarrow\sqrt{x+3}-\frac{5}{3}+\sqrt{4x+12}-\frac{10}{3}=0\)

\(\Leftrightarrow\frac{x+3-\frac{25}{9}}{\sqrt{x+3}+\frac{5}{3}}+\frac{4x+12-\frac{100}{9}}{\sqrt{4x+12}+\frac{10}{3}}=0\)

\(\Leftrightarrow\frac{\frac{9x+2}{9}}{\sqrt{x+3}+\frac{5}{3}}+\frac{\frac{4\left(9x+2\right)}{9}}{\sqrt{4x+12}+\frac{10}{3}}=0\)

\(\Leftrightarrow\frac{9x+2}{9}\left(\frac{1}{\sqrt{x+3}+\frac{5}{3}}+\frac{4}{\sqrt{4x+12}+\frac{10}{3}}\right)=0\)

Pt \(\frac{1}{\sqrt{x+3}+\frac{5}{3}}+\frac{4}{\sqrt{4x+12}+\frac{10}{3}}>0\forall x\ge3\)

\(\Rightarrow\frac{9x+2}{9}=0\Rightarrow9x+2=0\Rightarrow x=-\frac{2}{9}\)

Answer 2

[nguyễn hoàng tiến]

x = -2/9