b: Xét ΔMAN và ΔMPA có

góc MAN=góc MPA

góc AMN chung

=>ΔMAN đồng dạng với ΔMPA

=>MA/MP=MN/MA

=>MA^2=MN*MP

Với x >= 0 ; x khác 9 

\(B=\dfrac{x-3\sqrt{x}-x-6\sqrt{x}-9+x+11\sqrt{x}+6}{x-9}\)

\(=\dfrac{x+2\sqrt{x}-3}{x-9}=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{x-9}=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)

\(\Rightarrow P=A:B=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}:\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

\(\Delta=\left(-m\right)^2-4\left(2m-4\right)\)

\(=m^2-8m+16\)

\(=\left(m-4\right)^2>0\) khi \(m\ne4\)

\(x_1=\dfrac{m+m-4}{2}=m-2\)

\(x_2=\dfrac{m-m-4}{2}=-2\)

\(TH_1:\left(m-2\right)^2=5.\left(-2\right)-1\)

\(\Leftrightarrow\left(m-2\right)^2=-11\left(vô.lí\right)\)

\(TH_2:\left(-2\right)^2=5\left(m-2\right)-1\)

\(\Leftrightarrow5m=15\)

\(\Leftrightarrow m=3\left(TM\right)\)

Vậy ...

3) Ta có: \(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(m^2-6\right)\)

\(=\left(2m-2\right)^2-4\left(m^2-6\right)\)

\(=4m^2-8m+4-4m^2+24\)

\(=-8m+28\)

Để phương trình có hai nghiệm phân biệt x1;x2 thì Δ>0

\(\Leftrightarrow-8m+28>0\)

\(\Leftrightarrow-8m>-28\)

hay \(m< \dfrac{7}{2}\)

Áp dụng hệ thức Vi-et, ta được:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m-1\right)}{1}=2m-2\\x_1x_2=m^2-6\end{matrix}\right.\)

Ta có: \(x_1^2+x_2^2=16\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=16\)

\(\Leftrightarrow\left(2m-2\right)^2-2\left(m^2-6\right)-16=0\)

\(\Leftrightarrow4m^2-8m+4-2m^2+12-16=0\)

\(\Leftrightarrow2m^2-8m=0\)

\(\Leftrightarrow2m\left(m-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\left(nhận\right)\\m=4\left(loại\right)\end{matrix}\right.\)