a) ĐK: \(x\ge1\)
Ta có: \(\sqrt{4x-4}=\dfrac{x+3}{2}\)
<=> \(2\sqrt{4\left(x-1\right)}=x+3\)
<=> \(2.2\sqrt{x-1}=x+3\)
<=> \(x+3-4\sqrt{x-1}=0\)
<=> \(\left(x-1\right)-4\sqrt{x-1}+4=0\)
<=> \(\left(\sqrt{x-1}-2\right)^2=0\)
<=> \(\sqrt{x-1}=2\)
<=> \(x-1=4\) => \(x=5\) (TM)
Vậy ............................................
b) ĐK: \(x\ge1;y\ge2;z\ge3\)
Ta có: \(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
<=> \(\left(x-1\right)-2\sqrt{x-1}+1+\left(y-2\right)-4\sqrt{y-2}+4+\)
\(\left(z-3\right)-6\sqrt{z-3}+9=0\)
<=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}+3\right)^2=0\)
=> \(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\) (TM)
Vậy ............................................
