câu a dễ nên bạn tự giải nha( mk super lười)
b/ \(\sqrt{x-1}+\sqrt{x^2-1}=\sqrt{x^2-5x+4}\) ĐKXĐ:....
\(\Leftrightarrow\sqrt{x-1}+\sqrt{\left(x-1\right)\left(x+1\right)}=\sqrt{\left(x-\frac{5}{2}\right)^2-\frac{9}{4}}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{\left(x+1\right)\left(x-1\right)}=\sqrt{\left(x-\frac{5}{2}-\frac{3}{2}\right)\left(x-\frac{5}{2}+\frac{3}{2}\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{\left(x+1\right)\left(x-1\right)}=\sqrt{\left(x-4\right)\left(x-1\right)}\)
Đặt \(\sqrt{x-1}=a\ge0\Rightarrow\sqrt{x+1}=a+2;\sqrt{x-4}=a-3\)
\(\Rightarrow a+a\left(a+2\right)=a\left(a-3\right)\)
\(\Leftrightarrow a+a^2+2a=a^2-3a\)
\(\Leftrightarrow a=0\)
Thay trở lại có \(\sqrt{x-1}=0\Rightarrow x=1\)
