Question

Giải phương trình :

1) \(\sqrt{x+10}-\sqrt{3-4x}=2\sqrt{x+2}\)

2)\(\sqrt{1-\sqrt{x^4-x^2}}=x-1\)

Answer 1

1.

ĐKXĐ:\(-2\le x\le\frac{3}{4}\)

\(pt\Leftrightarrow\sqrt{x+10}=2\sqrt{x+2}+\sqrt{3-4x}\)

\(\Leftrightarrow x+10=4x+8+3-4x+2\sqrt{\left(4x+8\right)\left(3-4x\right)}\)

\(\Leftrightarrow2\sqrt{\left(4x+8\right)\left(3-4x\right)}=x-1\)

\(\Leftrightarrow2\sqrt{-4x^2-22x+24}=x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\4\left(-4x^2-22x+24\right)=\left(x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\17x^2-86x-95=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=\frac{43-2\sqrt{866}}{17}\left(l\right)\\x=\frac{43+2\sqrt{866}}{17}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm