a) ĐKXĐ: x≥-3
Ta có: \(\sqrt{x+3}+\frac{1}{2}\sqrt{9x+27}=10\)
\(\Leftrightarrow\sqrt{x+3}+\frac{1}{2}\cdot3\sqrt{x+3}=10\)
\(\Leftrightarrow\frac{5}{2}\cdot\sqrt{x+3}=10\)
\(\Leftrightarrow\sqrt{x+3}=4\)
\(\Leftrightarrow x+3=16\)
hay x=13(nhận)
Vậy: S={13}
b) Ta có: \(\sqrt{x^2-6x+9}=4\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=4\)
\(\Leftrightarrow\left|x-3\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
Vậy: S={7;-1}
c) ĐKXĐ: x≥3
Ta có: \(\sqrt{x^2-x-6}-2\sqrt{x-3}+\sqrt{x+2}-2=0\)
\(\Leftrightarrow\sqrt{x-3}\cdot\sqrt{x+2}-2\sqrt{x-3}+\sqrt{x+2}-2=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+2}-2\right)+\left(\sqrt{x+2}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+2}-2\right)\left(\sqrt{x-3}+1\right)=0\)
mà \(\sqrt{x-3}+1>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x+2}-2=0\)
\(\Leftrightarrow\sqrt{x+2}=2\)
\(\Leftrightarrow x+2=4\)
hay x=2(loại)
Vậy: S=∅
d) ĐKXĐ: x≥1
Ta có: \(x-4\sqrt{x-1}=6\)
\(\Leftrightarrow x=6+4\sqrt{x-1}\)
\(\Leftrightarrow x^2=36+16\left(x-1\right)+48\sqrt{x-1}\)
\(\Leftrightarrow x^2-36-16x+16=48\sqrt{x-1}\)
\(\Leftrightarrow x^2-16x-20=48\sqrt{x-1}\)
\(\Leftrightarrow x^4+256x^2+400-32x^3-40x^2+640x=2304\left(x-1\right)\)
\(\Leftrightarrow x^4-32x^3+216x^2+400+640x=2304x-2304\)
\(\Leftrightarrow x^4-32x^3+216x^2+640x+400-2304x+2304=0\)
\(\Leftrightarrow x^4-32x^3+216x^2-1664x+2704=0\)
\(\Leftrightarrow x^4-26x^3-6x^3+156x^2+60x^2-1560x-104x+2704=0\)
\(\Leftrightarrow x^3\left(x-26\right)-6x^2\left(x-26\right)+60x\left(x-26\right)-104\left(x-26\right)=0\)
\(\Leftrightarrow\left(x-26\right)\left(x^3-6x^2+60x-104\right)=0\)
\(\Leftrightarrow\left(x-26\right)\left(x^3-2x^2-4x^2+8x+52x-104\right)=0\)
\(\Leftrightarrow\left(x-26\right)\left[x^2\left(x-2\right)-4x\left(x-2\right)+52\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-26\right)\left(x-2\right)\left(x^2-4x+52\right)=0\)
mà \(x^2-4x+52>0\forall x\)
nên \(\left[{}\begin{matrix}x-26=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=26\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
Vậy: S={2;26}
