Question

Tìm x ( giải phương trình )

a. \(\sqrt{2}-1=\sqrt{5}-1\)

b. \(\sqrt{\left(4-3x^2\right)}=\sqrt{2}+1\)

c. \(\sqrt{x^2-9}+3\sqrt{x+3}=0\)

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

Answer 1

b: \(\sqrt{4-3x^2}=\sqrt{2}+1\)

\(\Leftrightarrow4-3x^2=3+2\sqrt{2}\)

\(\Leftrightarrow3x^2=1-2\sqrt{2}\)

\(\Leftrightarrow x^2=\dfrac{1-2\sqrt{2}}{3}< 0\)(vô lý)

c: \(\sqrt{x^2-9}+3\sqrt{x+3}=0\)

\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x-3}+3\right)=0\)

\(\Leftrightarrow\sqrt{x+3}=0\)

=>x=-3

d: \(\sqrt{x+2\sqrt{x}+1}=\sqrt{2}+1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=\sqrt{2}+1\\\sqrt{x}+1=-\sqrt{2}-1\end{matrix}\right.\Leftrightarrow\sqrt{x}=\sqrt{2}\)

hay x=2