a) \(x=\sqrt{2x+3}\) (đk \(x\ge-\dfrac{2}{3}\) )
\(\Leftrightarrow x^2-2x-3=0\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=3\left(nh\right)\end{matrix}\right.\)
b)ĐK \(x\le8\)
\(\Leftrightarrow x^2+x+12=\left(8-x\right)^2\)
\(\Leftrightarrow x^2+x+12=x^2-16x+64\)
\(\Leftrightarrow17x=52\Rightarrow x=\dfrac{52}{17}\)
c) ĐK \(x\ge1\)
\(\Leftrightarrow\left(2\sqrt{x-1}+\sqrt{x+2}\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow4\left(x-1\right)+x+2+4\sqrt{\left(x-1\right)\left(x+2\right)}=x^2+6x+9\)
\(\Leftrightarrow4\sqrt{x^2+x-2}=x^2+x+11\)
\(\Leftrightarrow=x^2+x+2-4\sqrt{x^2+x-2}+9=0\)( vô lí)
suy ra pt vô nghiệm
d) ĐK \(x\ge3\)
\(\Leftrightarrow x^2-6x+9-\left(x-3\right)-2\sqrt{x-3}+2=0\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)-2\sqrt{x-3}+2=0\)
Đặt \(t=\sqrt{x-3}\)
\(\Leftrightarrow t^4-t^2-2t+2=0\)
\(\Leftrightarrow t^2\left(t^2-1\right)-2\left(t-1\right)=0\)
\(\Leftrightarrow t^2\left(t-1\right)\left(t+1\right)-2\left(t-1\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(t^3+t^2-2\right)=0\)
\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t^2+2t+2=0\left(vl\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x-3}=1\Leftrightarrow x-3=1\Rightarrow x=4\)
