`c)` `x-9\sqrt{x}+14=0` Điều kiện: `x\geq0`
`<=>x-2\sqrt{x}-7\sqrt{x}+14=0`
`<=>(x-2\sqrt{x})-(7\sqrt{x}-14)=0`
`<=>\sqrt{x}(\sqrt{x}-2)-7(\sqrt{x}-2)=0`
`<=>(\sqrt{x}-7)(\sqrt{x}-2)=0`
`<=>` \(\left[ \begin{array}{l}\sqrt{x}-7=0\\\sqrt{x}-2=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}\sqrt{x}=7\\\sqrt{x}=2\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=49\quad(\text{nhận})\\x=4\quad(\text{nhận})\end{array} \right.\)
Vậy `x=49;x=4`
`d)` `\sqrt{x^2-10x+25}=7-7x` Điều kiện: `x\leq1`
`<=>\sqrt{(x-5)^2}=7-7x`
`<=>|x-5|=7-7x`
`<=>` \(\left[ \begin{array}{l}x-5=7-7x\\x-5=-(7-7x)\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x+7x=7+5\\x-5=-7+7x\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}8x=12\\-6x=-2\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\quad(\text{loại vì x > 1})\\x=\dfrac{1}{3}\quad(\text{nhận vì x < 1})\end{array} \right.\)
Vậy `x=1/3`
