a) \(\sqrt{3x+1}-\sqrt{x+4}=1\)
\(\Leftrightarrow4x+5-2\sqrt{\left(3x+1\right)\left(x+4\right)}=1\)
\(\Leftrightarrow-2\sqrt{\left(3x+1\right)\left(x+4\right)}=1-4x-5\)
\(\Leftrightarrow-2\sqrt{\left(3x+1\right)\left(x+4\right)}=-4x-4\)
Bình phương hai vế, ta có:
\(\Leftrightarrow4\left(3x+1\right)\left(x+4\right)=16\left(x+1\right)^2\)
\(\Leftrightarrow\left(3x+1\right)\left(x+4\right)=4\left(x+1\right)^2\)
\(\Leftrightarrow3x^2+12x+x+4=4x^2+8x+4\)
\(\Leftrightarrow3x^2+13x+4=4x^2+8x+4\)
\(\Leftrightarrow3x^2+13x=4x^2+8x\)
\(\Leftrightarrow3x^2+13x-4x^2+8x=0\)
\(\Leftrightarrow-x^2+5x=0\)
\(\Leftrightarrow5x-x^2=0\)
\(\Leftrightarrow x\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy nghiệm phương trình là {0; 5}
b) \(x-\sqrt{x}-6=0\)
\(\Leftrightarrow-\sqrt{x}=-x+6\)
\(\Leftrightarrow-\sqrt{x}=6-x\)
Bình phương hai vế, ta có:
\(\Leftrightarrow x=36-12x+x^2\)
\(\Leftrightarrow x-36+12x+x^2=0\)
\(\Leftrightarrow13x-36-x^2=0\)
\(\Leftrightarrow\left(x-9\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)
Vậy nghiệm phương trình là {9; 4}
c) \(x=\sqrt{10-x}-2\)
\(\Leftrightarrow x+2=\sqrt{10-x}\)
Bình phương hai vế, ta có:
\(\Leftrightarrow x^4+4x+4=10-x\)
\(\Leftrightarrow x^2+4x+4-10+x=0\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\)
Vậy nghiệm phương trình là {1; -6}
