Giải thích các bước giải:
b.Từ phương trình ta có:
$\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac1{18}$
$\to \dfrac{(x+5)-(x+4)}{(x+4)(x+5)}+\dfrac{(x+6)-(x+5)}{(x+5)(x+6)}+\dfrac{(x+7)-(x+6)}{(x+6)(x+7)}=\dfrac1{18}$
$\to \dfrac1{x+4}-\dfrac1{x+5}+\dfrac1{x+5}-\dfrac1{x+6}+\dfrac1{x+6}-\dfrac1{x+7}=\dfrac1{18}$
$\to \dfrac1{x+4}-\dfrac1{x+7}=\dfrac1{18}$
$\to 18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)$
$\to 54=x^2+11x+28$
$\to x^2+11x-26=0$
$\to (x-2)(x+13)=0$
$\to x\in\{2,-13\}$
c.Từ phương trình suy ra:
$(9x^2-18x+9)+(y^2-6y+9)+(2z^2+4z+2)=0$
$\to 9(x-1)^2+(y-3)^2+2(z+2)^2=0$
$\to x-1=y-3=z+2=0$
$\to x=1, y=3,z=-2$
