Giải phương trình: [tex]\sqrt{2x+2\sqrt{2x-5}-4}+\sqrt{2x-6\sqrt{2x-5}+4}=4[/tex]
ĐKXĐ: [tex]x\geq \frac{5}{2}[/tex]
[tex]\sqrt{2x+2\sqrt{2x-5}-4}+\sqrt{2x-6\sqrt{2x-5}+4}=4[/tex]
[tex]\Leftrightarrow \sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5-6\sqrt{2x-5}+9}=4[/tex]
[tex]\Leftrightarrow \sqrt{(\sqrt{2x-5}+1)^2}+\sqrt{(\sqrt{2x-5}-3)^2}=4[/tex]
[tex]\Leftrightarrow |\sqrt{2x-5}+1|+|\sqrt{2x-5}-3|=4[/tex]
[tex]\Leftrightarrow \sqrt{2x-5}+1+|\sqrt{2x-5}-3|=4[/tex]
Có: [tex]|\sqrt{2x-5}-3|=|3-\sqrt{2x-5}|\geq 3-\sqrt{2x-5}[/tex] với mọi x
=> [tex]\sqrt{2x-5}+1+|\sqrt{2x-5}-3|\geq \sqrt{2x-5}+1+3-\sqrt{2x-5}=4[/tex]
Dấu "=" xảy ra [tex]\Leftrightarrow |\sqrt{2x-5}-3| = 3-\sqrt{2x-5} \Leftrightarrow \sqrt{2x-5}-3 \leq 0 \Leftrightarrow \sqrt{2x-5} \leq 3 \Leftrightarrow 2x-5 \leq 9 \Leftrightarrow x \leq 7 [/tex] KH ĐKXĐ
=> [tex]\frac{5}{2}\leq x\leq 7[/tex]
Vậy....