Question

sa se rezolve ecuatiile :√2x-1+√2-x=2

Answer 1

[tex]\text{Mai intai stabilim conditiile de existenta ale radicalului:}\\ 2x-1\geq0 \Rightarrow x\geq \dfrac{1}{2}\\ 2-x\geq 0\Rightarrow x\leq 2\\ C.E.:x\in \left[\dfrac{1}{2},2\right]\\ \sqrt{2x-1}+\sqrt{2-x}=2|()^2\\ 2x-1+2-x+2\sqrt{(2x-1)(2-x)}=4\\ x+1+2\sqrt{(2x-1)(2-x)}=4\\ 2\sqrt{(2x-1)(2-x)}=3-x|()^2\\ 4(2x-1)(2-x)=9-6x+x^2\\ 4(4x+x-2-2x^2)=9-6x+x^2\\ 20x-8-8x^2=9-6x+x^2\\ 9x^2-26x+17=0\\ 9x^2-9x-17x+17=0\\ 9x(x-1)-17(x-1)=0\\ (x-1)(9x-17)=0\Rightarrow x\in \left\{1,\dfrac{17}{9}\right\}[/tex]