Đáp án:
Giải thích các bước giải:
5. `13\sqrt{x-1}+9\sqrt{x+1}=16x`
ĐK: `x \ge 1`
`⇔ 16-13\sqrt{x-1}-9\sqrt{x+1}=0`
`⇔ 13(x-1-\sqrt{x-1}+\frac{1}{4})+3(x+1-3\sqrt{x+1}+\frac{9}{4})=0`
`⇔ 13[(\sqrt{x-1})^2-2\sqrt{x-1}.\frac{1}{2}+(\frac{1}{2})^2]+3[(\sqrt{x+1})^2-2\sqrt{x+1}.\frac{3}{2}+(\frac{3}{2})^2]=0`
`⇔ 13(\sqrt{x-1}-\frac{1}{2})^2+3(\sqrt{x+1}-\frac{3}{2})^2=0`
`⇔` \(\left[ \begin{array}{l}\sqrt{x-1}-\dfrac{1}{2}=0\\\sqrt{x+1}-\dfrac{3}{2}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{5}{4}\\x=\dfrac{5}{4}\end{array} \right.\)
`⇔ x=5/4\ (TM)`
Vậy `S={5/4}`
