a, M=\(\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)(ĐKXD: x>0, x#4, x#9)
=\(\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)^{ }}\)=\(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
Vậy.....
b, ta có x=11-6\(\sqrt{2}\)=\(\left(3-\sqrt{2}\right)^2\)
Thay vào M ta đươc:
M=\(\frac{\sqrt{\left(3-\sqrt{2}\right)^2}+1}{\sqrt{\left(3-\sqrt{2}\right)^2}-3}\)=\(\frac{3-\sqrt{2}+1}{3-\sqrt{2}-3}=\frac{4-\sqrt{2}}{-\sqrt{2}}=1-2\sqrt{2}\)
c,Để M<1<=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)<1 <=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)-1<0<=> \(\frac{4}{\sqrt{x}-3}\)<0<=> x<9(t/m x#9) mà x>0, x#4 => 0<x<9 và x#4
Vậy....
d, Để M∈Z <=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)∈Z<=>\(1+\frac{4}{\sqrt{x}-3}\)∈Z<=>\(\frac{4}{\sqrt{x}-3}\)∈Z<=> 4⋮\(\sqrt{x}-3\)<=>\(\sqrt{x}-3\)∈Ư(4)={\(\pm\)1,\(\pm\)2,\(\pm\)4}
<=>\(\sqrt{x}\) ∈ {2,4,5,1,7}
<=>x ∈ {4,16,25,1,49} mà x#4
=> x∈ {16,25,1,49}
vậy..