Giải thích các bước giải:
Bài 7:
Ta có :
$\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=2$
$\dfrac{2b+c}{a}-1=\dfrac{2c+a}{b}-1=\dfrac{2a+b}{c}-1$
$\dfrac{2b+c}{a}=\dfrac{2c+a}{b}=\dfrac{2a+b}{c}=\dfrac{2b+c+2c+a+2a+b}{a+b+c}=3$
$\to \begin{cases}2b+c=3a\\ 2c+a=3b\\2a+b=3c\end{cases}$
$\to \begin{cases}2b+c=3a\\ a=3b-2c\\2a=3c-b\end{cases}$
$\to 2(3b-2c)=2a=3c-b\to b=c\to a=3b-2b=b\to a=b=c$
$\to M=\dfrac{(3a-2b)(3b-2c)(3c-2a)}{(3a-c)(3b-a)(3c-b)}=\dfrac{(3a-2a)(3a-2a)(3a-2a)}{(3a-a)(3a-a)(3a-a)}=\dfrac{a^3}{8a^3}=\dfrac18$
Bài 8:
Ta có : $(2x-4)^6\ge 0\quad\forall x$
$(y-7)^{12}\ge 0\quad\forall y$
$\to Q\ge 0+0-21=-21$
Dấu = xảy ra khi $x=2, y=7$
