Question

Giải phương trình :

a.\(\sqrt{x-1+2\sqrt{ }x-2}+\sqrt{x-1-2\sqrt{ }x-2}=1\)

Answer 1

A=\(\sqrt{x-1+2\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}\)

=\(\sqrt{x-2+2\sqrt{x-2}+1}+\sqrt{x-2-2\sqrt{x-2}+1}\)

=\(\sqrt{\left(\sqrt{x-2}+1\right)^2}+\sqrt{\left(\sqrt{x-2}-1\right)^2}\)

=\(\left|\sqrt{x-2}+1\right|+\left|\sqrt{x-2}-1\right|\)

=\(\sqrt{x-2}+1+\left|\sqrt{x-2}-1\right|\)

A=1 <=>\(\sqrt{x-2}+1+\left|\sqrt{x-2}-1\right|\)=1

<=>\(\sqrt{x-2}+\left|\sqrt{x-2}-1\right|\)=0

mà \(\left\{{}\begin{matrix}\sqrt{x-2}\ge0\\\left|\sqrt{x-2}-1\right|\ge0\end{matrix}\right.\)

=>dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x-2}-1=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

=>vô lí

=>pt vô nghiệm