Iracionaliosios lygties sprendinys

Išspręskite  lygtį :    √(x+1/2+√(x+1/4))=(x+1/4)²+1/2 (3t)

Panaudokite keitinį

[tex]t=\sqrt{x+\frac{1}{4}}, \space t≥0[/tex]
[tex]x+\frac{1}{4}=t^2\implies x=t^2-\frac{1}{4}[/tex]
Kai [tex]t≥0:[/tex]
[tex]\sqrt{t^2-\frac{1}{4}+\frac{1}{2}+\sqrt{t^2}}=(t^2)^2+\frac{1}{2}\implies \sqrt{t^2+\frac{1}{4}+|t|}=t^4+\frac{1}{2}\implies \\\sqrt{t^2+2t\cdot \frac{1}{2}+(\frac{1}{2})^2}=t^4+\frac{1}{2}\implies \sqrt{(t+\frac{1}{2})^2}=t^4+\frac{1}{2}\implies |t+\frac{1}{2}|=t^4+\frac{1}{2}\implies\\ t+\frac{1}{2}=t^4+\frac{1}{2}\implies t-t^4=0\implies t(1-t^3)=0\implies t=0,\space 1-t^3=0\implies t=0,\space t=1.[/tex]
[tex]t=0\implies x=0^2-\frac{1}{4}=-\frac{1}{4}[/tex]
[tex]t=1\implies x=1^2-\frac{1}{4}=\frac{3}{4}[/tex]
Ats.: [tex]-\frac{1}{4};\frac{3}{4}[/tex]

Taip. Aš gaunu tokius pat atsakymus -1/4 ir 3/4.

Teisingai Galima trumpiau  √((x+1/4)+√(x+1/4)+1/4)=√(√(x+1/4)+1/2)²= √(x+1/4)+1/2              √(x+1/4)+1/2=(x+1/4)²+1/2    √(x+1/4)=(x+1/4)²      x+1/4=t      √t=t²...........