Giải thích các bước giải:
1.Ta có:
$x^2+2x+y^2-4y+5=0$
$\to (x^2+2x+1)+(y^2-4y+4)=0$
$\to (x+1)^2+(y-2)^2=0$
Mà $ (x+1)^2+(y-2)^2\ge 0$
$\to$Dấu = xảy ra khi $x+1=y-2=0\to x=-1, y=2$
2.Ta có:
$\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}$
$\to (\dfrac{8}{x-8}+1)+(\dfrac{11}{x-11}+1)=(\dfrac{9}{x-9}+1)+(\dfrac{10}{x-10}+1)$
$\to \dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}$
$\to \dfrac{x}{x-8}+\dfrac{x}{x-11}=\dfrac{x}{x-9}+\dfrac{x}{x-10}$
$\to x(\dfrac{1}{x-8}+\dfrac{1}{x-11})=x(\dfrac{1}{x-9}+\dfrac{1}{x-10})$
$\to x(\dfrac{1}{x-8}+\dfrac{1}{x-11})-x(\dfrac{1}{x-9}+\dfrac{1}{x-10})=0$
$\to x(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10})=0$
$\to x=0$
Hoặc $\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0$
$\to \dfrac{1}{x-8}-\dfrac{1}{x-9}=\dfrac{1}{x-10}-\dfrac{1}{x-11}$
$\to \dfrac{(x-9)-(x-8)}{(x-8)(x-9)}=\dfrac{(x-11)- (x-10)}{(x-10)(x-11)}$
$\to \dfrac{-1}{(x-8)(x-9)}=\dfrac{-1}{(x-10)(x-11)}$
$\to (x-8)(x-9)=(x-10)(x-11)$
$\to x^2-17x+72=x^2-21x+110$
$\to 4x=38$
$\to x=\dfrac{19}{2}$
Vậy $x\in\{0,\dfrac{19}{2}\}$
c.Ta có:
$\dfrac{1}{x-1}+\dfrac{2}{x-2}+\dfrac{3}{x-3}=\dfrac{6}{x-6}$
$\to 1(x-2)(x-3)(x-6)+2(x-1)(x-3)(x-6)+3(x-1)(x-2)(x-6)=6(x-1)(x-2)(x-3)$
$\to 6x^3-58x^2+150x-108=6x^3-36x^2+66x-36$
$\to -22x^2+84x-72=0$
$\to x=\dfrac{21\pm3\sqrt{5}}{11}$
