Mọi người giúp em vs ạ Giải pt: a) $x./x-1/=-(x-1)^{2}$ ( //: giá trị tuyệt đối) b) $\frac{x^{2}-3x-m^{2}}{\sqrt[]{x-m^{2}}}$ =

Đáp án:

$\begin{array}{l}
a)x.\left| {x – 1} \right| =  – {\left( {x – 1} \right)^2}\left( {x \le 0} \right)\\
 \Rightarrow x\left( {1 – x} \right) + {\left( {x – 1} \right)^2} = 0\\
 \Rightarrow \left( {x – 1} \right).\left( {x – 1 – x} \right) = 0\\
 \Rightarrow \left( {x – 1} \right).\left( { – 1} \right) = 0\\
 \Rightarrow x = 1\left( {ktm} \right)\\
 \Rightarrow x \in \emptyset \\
b)\frac{{{x^2} – 3x – {m^2}}}{{\sqrt {x – {m^2}} }} = \sqrt {x – {m^2}} \\
 \Rightarrow {x^2} – 3x – {m^2} = x – {m^2}\\
 \Rightarrow {x^2} – 4x = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 4
\end{array} \right.\\
c)Dkxd:{x^2} \le 17\\
x + \sqrt {17 – {x^2}}  = a\\
 \Rightarrow 17 + 2x\sqrt {17 – {x^2}}  = {a^2}\\
 \Rightarrow x\sqrt {17 – {x^2}}  = \frac{{{a^2} – 17}}{2}\\
pt \Rightarrow a + \frac{{{a^2} – 17}}{2} = 9\\
 \Rightarrow {a^2} + 2a – 17 – 18 = 0\\
 \Rightarrow {a^2} + 2a – 35 = 0\\
 \Rightarrow \left[ \begin{array}{l}
a =  – 7\\
a = 5
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x + \sqrt {17 – {x^2}}  =  – 7\\
x + \sqrt {17 – {x^2}}  = 5
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\sqrt {17 – {x^2}}  =  – 7 – x\left( {ktm} \right)\\
\sqrt {17 – {x^2}}  = 5 – x\left( {x \le 5} \right)
\end{array} \right.\\
 \Rightarrow 17 – {x^2} = {x^2} – 10x + 25\\
 \Rightarrow 2{x^2} – 10x + 8 = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 4\left( {tm} \right)\\
x = 1\left( {tm} \right)
\end{array} \right.
\end{array}$