`#Rùa`
Đáp án+Giải thích các bước giải:
`a)\sqrt{4x-16}+\sqrt{x-4}-1/3 \sqrt{9x-36}=4` (`x ≥4`)
`<=> \sqrt{4(x-4)}+\sqrt{x-4}-1/3 \sqrt{9(x-4)}=4`
`<=> 2\sqrt{x-4}+\sqrt{x-4}-\sqrt{x-4}=4`
`<=> 2\sqrt{x-4}=4`
`<=> \sqrt{x-4}=2`
`<=> x-4=4`
`<=> x=8` (TM)
Vậy `S={8}`
`b)\sqrt{9x-9}-\sqrt{4x-4}+\sqrt{16x-16}-3\sqrt{x-1}=16` (`x ≥1`)
`<=> \sqrt{9(x-1)}-\sqrt{4(x-1)}+\sqrt{16(x-1)}-3\sqrt{x-1}=16`
`<=> 3\sqrt{x-1}-2\sqrt{x-1}+4\sqrt{x-1}-3\sqrt{x-1}=16`
`<=> 2\sqrt{x-1}=16`
`<=> \sqrt{x-1}=8`
`<=> x-1=64`
`<=> x=65` (TM)
Vậy `S={65}`
`c) 2x-7\sqrt{x}+5=0` (`x≥0`)
`<=> (2x-2\sqrt{x})-(5\sqrt{x}-5)=0`
`<=> 2\sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0`
`<=> (\sqrt{x}-1)(2\sqrt{x}-5)=0`
`<=>` \(\left[ \begin{array}{l}\sqrt{x}-1=0\\2\sqrt{x}-5=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}\sqrt{x}=1\\2\sqrt{x}=5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=1(TM)\\x=\dfrac{25}{4}(TM)\end{array} \right.\)
Vậy `S={1;25/4}`
`d) x-6\sqrt{x-3}-10=0` (`x≥3`)
`<=> [(x-3)-6\sqrt{x-3}+9]-16=0`
`<=> (\sqrt{x-3}-3)^2=16`
`<=> \sqrt{x-3}-3=±4`
`<=>` \(\left[ \begin{array}{l}\sqrt{x-3}-3=4\\\sqrt{x-3}-3=-4\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}\sqrt{x-3}=7\\\sqrt{x-3}=-1\end{array} \right.\)
`<=> \sqrt{x-3}=7 ` (Vì `\sqrt{x-3}≥0`)
`<=> x-3=49`
`<=> x=52` (TM)
Vậy `S={52}`
`e)\sqrt{3x+2}=\sqrt{5x-4}` (`x≥5/4`)
`<=> 3x+2=5x-4`
`<=> 2x=6`
`<=> x=3` (TM)
Vậy `S={3}`
`f)\sqrt{4x+1}=3x+1` (`x ≥ -1/4`)
`<=> 4x+1=(3x+1)^2`
`<=> 4x+1=9x^2+6x+1`
`<=> 9x^2+2x=0`
`<=> x(9x+2)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\9x+2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=0(TM)\\x=\dfrac{-2}{9}(TM)\end{array} \right.\)
Vậy `S={0;-2/9}`
`g) (x+3)\sqrt{x-1}=0` (`x ≥3`)
`<=>` \(\left[ \begin{array}{l}x+3=0\\\sqrt{x-1}=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-3(koTM)\\x=1(TM)\end{array} \right.\)
Vậy `S={1}`
