a, \(x^2-3x-6+4=0\)

\(\Leftrightarrow x^2-3x-2=0\)

Ta có : \(\left(-3\right)^2-4.\left(-2\right)=9+8=17>0\)

Nên có 2 nghiệm phân biệt 

\(x_1=\frac{3-\sqrt{17}}{2};x_2=\frac{3+\sqrt{17}}{2}\)

b, Để PT có nghiệm thì \(\Delta=0\)

\(\Leftrightarrow b^2-4ac=0\)

\(\Leftrightarrow\left(-3\right)^2-4\left(-m+4\right)=0\)

\(\Leftrightarrow9+4m-16=0\)

\(\Leftrightarrow7+4m=0\)

\(\Leftrightarrow m=-\frac{7}{4}\)

Vậy => m = -7/4 

c, Ko rõ 

Xét \(\Delta'=m^2-4m+4=\left(m-2\right)^2\ge0\) luôn có nghiệm

Theo hệ thức Viete ta có:

\(x_1+x_2=2m;x_1x_2=4m-4\Rightarrow x_1=2m-x_2\)

\(\Leftrightarrow2x_2=2m-x_2\Leftrightarrow3x_2=2m\Leftrightarrow x_2=\frac{2m}{3}\Rightarrow x_1=\frac{4m}{3}\)

\(\Rightarrow\frac{2m}{3}\cdot\frac{4m}{3}=4m-4\Leftrightarrow8m^2-36m+36=0\)

\(\Leftrightarrow m=3;m=\frac{3}{2}\)

Vậy m=3;m=³/₂ 

Không chắc lắm đâu nha !

\(P=\frac{3x-6\sqrt{x}+7}{2\sqrt{x}-2}+\frac{y-4\sqrt{x}+10}{\sqrt{y}-2}\)

\(=\frac{3\left(\sqrt{x}-1\right)}{2}+\frac{4}{2\left(\sqrt{x}-1\right)}+\left(\sqrt{y}-2\right)+\frac{6}{\sqrt{y-1}}\)

\(=\frac{3\left(\sqrt{x}-1\right)}{2}+\frac{3}{2\left(\sqrt{x}-1\right)}+\left(\sqrt{y}-2\right)+\frac{4}{\left(\sqrt{y}-2\right)}+\frac{4}{2\left(\sqrt{y}-2\right)}+\frac{1}{2\left(\sqrt{x}-1\right)}\)

\(\ge2.\sqrt{\frac{3}{2}.\frac{3}{2}}+2\sqrt{4}+\frac{\left(1+2\right)^2}{2\left(\sqrt{x}+\sqrt{y}-3\right)}\)

\(=3+4+\frac{3}{2}=\frac{17}{2}\)

Dấu "=" xảy ra <=> x = 4 và y = 16