\(\dfrac{\sqrt{5x+7}}{\sqrt{x+3}}=4\left(x\ge-\dfrac{7}{5};x\ne-3\right)\)
\(< =>\sqrt{5x+7}=4\sqrt{x+3}\)
\(< =>5x+7=16\left(x+3\right)\)
`<=>5x+7=16x+48`
`<=>5x-16x=48-7`
`<=>-11x=41`
`<=>x=-41/11(ktm)`
Vậy pt vô nghiệm
\(\dfrac{\sqrt{5x+7}}{\sqrt{x+3}}=4\) ĐK: \(x\ge-\dfrac{7}{5}\)
\(\Rightarrow\sqrt{5x+7}=4\sqrt{x+3}\)
\(\Leftrightarrow\sqrt{\left(5x+7\right)^2}=4^2\sqrt{\left(x+3\right)^2}\)
\(\Leftrightarrow5x+7=16\left(x+3\right)\)
\(\Leftrightarrow5x+7=16x+48\)
\(\Leftrightarrow16x-5x=7-48\)
\(\Leftrightarrow11x=-41\)
\(\Leftrightarrow x=-\dfrac{41}{11}\) (loại)
Vậy \(S=\varnothing\).
