\(\sqrt{3x-2}+\sqrt{x+3}=5\) ( ĐK: \(x\ge\frac{2}{3}\) )
\(\Leftrightarrow3x-2+x+3+2\sqrt{\left(3x-2\right)\left(x+3\right)}=25\)
\(\Leftrightarrow2\sqrt{\left(3x-2\right)\left(x+3\right)}=24-4x\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(3x-2\right)\left(x+3\right)=\left(24-4x\right)^2\\24-4x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-55x-600=0\left(1\right)\\x\le6\end{matrix}\right.\)
Giải pt (1), ta được
\(x_1=\frac{55-5\sqrt{97}}{2}\left(tm\right)\)
\(x_2=\frac{55-5\sqrt{97}}{2}\left(tm\right)\)
P/s : Có gì sai sót mong bn thông cảm ạ
=.= hk tốt!!
\(2\sqrt{x-1}-\sqrt{x+1}=3\) (ĐK: x\(x\ge1\))
\(\Leftrightarrow2x-2+x+1-4\sqrt{\left(x+1\right)\left(x-1\right)}=9\)
\(\Leftrightarrow4\sqrt{x^2-1}=3x-10\)
\(\Leftrightarrow\left\{{}\begin{matrix}16\left(x^2-1\right)=\left(3x-10\right)^2\\3x-10\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x^2+60x-116=0\left(1\right)\\x\ge\frac{10}{3}\end{matrix}\right.\)
Giải phương trình (1), ta đc:
\(x_1=\frac{-30+4\sqrt{107}}{7}\left(ktm\right)\)
\(x_2=\frac{-30-4\sqrt{107}}{7}\left(ktm\right)\)
=> pt vô nghiệm
P/s: Có j sai sót thì bn thông cảm giùm mk nha :v
=.= hk tốt!!