Làm giúp mk câu 22 với

Giải thích các bước giải:

a.Ta có:

$a+b+c=0$

$\to b+c=-a$

$\to (b+c)^2=(-a)^2$

$\to (b+c)^2-4bc=(-a)^2-4bc$

$\to (b-c)^2=a^2-4bc$

$\to 4bc-a^2=-(b-c)^2$

Tương tự:

$4ca-b^2=-(c-a)^2$

$4ab-c^2=(a-b)^2$

Ta có:

$bc+2a^2=b(-a-b)+2a^2=-ab-b^2+2a^2=(2a^2-2ab)+(ab-b^2)=2a(a-b)+b(a-b)=(2a+b)(a-b)=(a+b+c+a-c)(a-b)=(a-c)(a-b)=-(a-b)(c-a)$

Tương tự:

$ca+2b^2=-(b-c)(c-a)$

$ab+2c^2=-(c-a)(b-c)$

$\to M=\dfrac{-(b-c)^2}{-(a-b)(c-a)}\cdot \dfrac{-(c-a)^2}{-(b-c)(c-a)}\cdot \dfrac{-(a-b)^2}{-(b-c)(c-a)}$

$\to M=1$

b.Ta có:

$N=(1+\dfrac ab)(1+\dfrac bc)(1+\dfrac ca)$

$\to N=\dfrac{b+a}b\cdot \dfrac{c+b}c\cdot \dfrac{a+c}a$

$\to N=\dfrac{-c}b\cdot \dfrac{-a}c\cdot \dfrac{-b}a$

$\to N=-1$