\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\) (*) (đk : \(x\ge\frac{5}{2}\))
Đặt \(\sqrt{2x-5}=a\left(a\ge0\right)\)
=> 2x-5=a2
<=> \(x=\frac{a^2+5}{2}\)
Có \(\sqrt{\frac{a^2+5}{2}-2+a}+\sqrt{\frac{a^2+5}{2}+2+3a}=7\sqrt{2}\)
<=> \(\sqrt{\frac{a^2+5-4+2a}{2}}+\sqrt{\frac{a^2+5+4+6a}{2}}=7\sqrt{2}\)
<=>\(\sqrt{\frac{a^2+2a+1}{2}}+\sqrt{\frac{a^2+6a+9}{2}}=7\sqrt{2}\)
<=> \(\frac{\sqrt{\left(a+1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(a+3\right)^2}}{\sqrt{2}}=7\sqrt{2}\)
<=> \(\left|a+1\right|+\left|a+3\right|=7\sqrt{2}.\sqrt{2}\)
<=> \(a+1+a+3=14\)(do a\(\ge\)0)
<=> \(2a=10\) <=> a=5(t/m)
<=> \(\sqrt{2x-5}=5\)
<=> \(2x-5=25\) <=> \(x=15\)(tm pt (*))
Vậy pt (*) có tập nghiệm \(S=\left\{15\right\}\)
