\(\frac{5}{\sqrt{x^2}+1}\)hay\(\frac{5}{\sqrt{x^2+1}}\)v
b)
Đặt \(\sqrt{x-2}=a\); \(\sqrt{4-x}=b\)
Ta có hpt:
\(\hept{\begin{cases}a+b=-a^2b^2+3\\a^2+b^2=2\end{cases}\Leftrightarrow\hept{\begin{cases}a+b=-a^2b^2+3\\\left(a+b\right)^2-2ab-2=0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=2\\\left(-a^2b^2+3\right)^2-2ab-2=0\end{cases}}\)
Đặt ab=t rồi giải hệ nhé bạn
Phần b cách ngắn hơn nè:
\(\sqrt{x-2}-1+\sqrt{4-x}-1=x^2-6x+9\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2}\right)^2-1}{\sqrt{x-2}+1}+\frac{\left(\sqrt{4-x}\right)^2-1}{\sqrt{4-x}+1}=\left(x-3\right)^2\)
\(\Leftrightarrow\frac{x-3}{\sqrt{x-2}+1}+\frac{3-x}{\sqrt{4-x}+1}=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}-\frac{1}{\sqrt{4-x}+1}-x+3\right)=0\)
\(\Rightarrow x=3\)
\(\sqrt{x^2-1}+\sqrt{x^2-9}=8\)
\(\Leftrightarrow\sqrt{\sqrt{x^2-8}+\sqrt{x^2-9}}+\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{7x+2\sqrt{x}}+9\sqrt{x^2-9}=0\)
\(\Leftrightarrow7x\sqrt{x^2+8}+2\sqrt{x^2-6}=6\)
\(\Leftrightarrow x^2-5=0\)
\(\Leftrightarrow x=\pm\sqrt{5}\)
