Đáp án + giải thích các bước giải:
3/ `\sqrt{3+\sqrt{x-4}}=3 (x\ge4)`
`->3+\sqrt{x-4}=9`
`->\sqrt{x-4}=6`
`->x-4=36`
`->x=40(TM)`
4/ `\sqrt{(x+3)^2-12x}=2x-1 (x\ge1/2)`
`->\sqrt{x^2+6x+9-12x}=2x-1`
`->\sqrt{x^2-6x+9}=2x-1`
`->\sqrt{(x-3)^2}=2x-1`
`->|x-3|=2x-1`
`->`\(\left[ \begin{array}{l}x-3=2x-1\\x-3=1-2x\end{array} \right.\)
`->`\(\left[ \begin{array}{l}-2=x(KTM)\\3x=4\end{array} \right.\)
`->x=4/3(TM)`
6/ `\sqrt{4x^2-4x+1}-2=x (x\ge-2)`
`->\sqrt{(2x-1)^2}=x+2`
`->|2x-1|=x+2`
`->`\(\left[ \begin{array}{l}2x-1=x+2\\1-2x=x+2\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=3\\-3x=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=3(TM)\\x=-\dfrac{1}{3}(TM)\end{array} \right.\)
7/ `\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}`
`->\sqrt{(3x-1)^2}=\sqrt{(x+4)^2}`
`->|3x-1|=|x+4|`
`->`\(\left[ \begin{array}{l}3x-1=x+4\\1-3x=x+4\end{array} \right.\)
`->`\(\left[ \begin{array}{l}2x=5\\-3=4x\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{array} \right.\)
10/ `\sqrt{x^2-25}-\sqrt{x-5}=0 (x>=5)`
`->\sqrt{(x-5)(x+5)}-\sqrt{x-5}=0`
`->\sqrt{x-5}(\sqrt{x+5}-1)=0`
`->`\(\left[ \begin{array}{l}\sqrt{x-5}=0\\\sqrt{x+5}-1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x-5=0\\x+5=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=5(TM)\\x=-4(KTM)\end{array} \right.\)
