Question

Giải phương trình:

a) \(\sqrt[3]{1-2x}+3=0\)

b) \(\sqrt{x-4\sqrt{x}+4}+\sqrt{x+6\sqrt{x}+9}=5\)

Answer 1

\(a.\sqrt[3]{1-2x}+3=0\)

\(\Leftrightarrow1-2x=-27\)

\(\Leftrightarrow x=14\)

b. \(\sqrt{x-4\sqrt{x}+4}+\sqrt{x+6\sqrt{x}+9}=5\left(ĐK:x\ge2\right)\)

\(\Leftrightarrow\sqrt{x-2.\sqrt{x}.2+2^2}+\sqrt{x+2.\sqrt{x}.3+3^2}=5\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}+3\right)^2}=5\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2+\sqrt{x}+3=5\\2-\sqrt{x}+\sqrt{x}+3=5\end{matrix}\right.\)

\(\Leftrightarrow x=4\left(n\right)\)

Lâu rồi mới trở lại :)))