giải pt: \(5\sqrt{x-1}+9\sqrt{x+1}=8x+6\) 31/07/2021 Bởi Melody giải pt: \(5\sqrt{x-1}+9\sqrt{x+1}=8x+6\)
Ta có: \(5\sqrt{x-1}+9\sqrt{x+1}=10\sqrt{\frac{1}{4}(x-1)}+6\sqrt{\frac{9}{4}(x+1)}\) Áp dụng BĐT Am-Gm ta có: \(\sqrt{\frac{1}{4}(x-1)}\leq \frac{x-1+\frac{1}{4}}{2}\) \(\sqrt{\frac{9}{4}(x+1)}\leq \frac{\frac{9}{4}+x+1}{2}\) Do đó, \(5\sqrt{x-1}+9\sqrt{x+1}\leq 5(x-1+\frac{1}{4})+3(\frac{9}{4}+x+1)\) \(\Leftrightarrow 5\sqrt{x-1}+9\sqrt{x+1}\leq 8x+6\) Dấu bằng xảy ra khi \(\left\{\begin{matrix} x-1=\frac{1}{4}\\ x+1=\frac{9}{4}\end {matrix}\right.\Leftrightarrow x=\frac{5}{4}\) Bình luận
Ta có: \(5\sqrt{x-1}+9\sqrt{x+1}=10\sqrt{\frac{1}{4}(x-1)}+6\sqrt{\frac{9}{4}(x+1)}\)
Áp dụng BĐT Am-Gm ta có:
\(\sqrt{\frac{1}{4}(x-1)}\leq \frac{x-1+\frac{1}{4}}{2}\)
\(\sqrt{\frac{9}{4}(x+1)}\leq \frac{\frac{9}{4}+x+1}{2}\)
Do đó, \(5\sqrt{x-1}+9\sqrt{x+1}\leq 5(x-1+\frac{1}{4})+3(\frac{9}{4}+x+1)\)
\(\Leftrightarrow 5\sqrt{x-1}+9\sqrt{x+1}\leq 8x+6\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x-1=\frac{1}{4}\\ x+1=\frac{9}{4}\end {matrix}\right.\Leftrightarrow x=\frac{5}{4}\)