Bài 2

$M=a(a+b)(a+c)$

$N=b(b+c)(b+a)$

$P=c(c+a)(c+b)$

Biết:

$a+b+c=0$

Ta có:

$a+b=-c$

$a+c=-b$

$b+c=-a$

Do đó:

$M=a(-c)(-b)$

$M=abc$

Tương tự:

$N=b(-a)(-c)$

$N=abc$

$P=c(-b)(-a)$

$P=abc$

Suy ra:

$\boxed{M=N=P=abc}$

Vậy $\boxed{M=N=P}$.

Điều kiện:

$3x+7\ge0\Rightarrow x\ge-\frac73$

Xét $|3x-2|$.

TH1: $x\ge\frac23$

$|3x-2|=3x-2$

$|2x+5-(3x-2)|=3x+7$

$|7-x|=3x+7$

Vì $x\ge\frac23$:

$7-x=3x+7$

$x=0$ (không thỏa $x\ge\frac23$).

$x-7=3x+7$

$x=-7$ (không thỏa).

Vô nghiệm.

TH2: $-\frac73\le x<\frac23$

$|3x-2|=2-3x$

$|2x+5-(2-3x)|=3x+7$

$|5x+3|=3x+7$

$5x+3=3x+7$

$x=2$ (không thuộc khoảng).

$-5x-3=3x+7$

$-8x=10$

$x=-\frac54$.

Thỏa mãn.

Vậy $\boxed{x=-\frac54}$.

Ta có:

$x[3|x+1|-2|x+2|]=12$

TH1: $x<-2$

$3x[-(x+1)]-2x[-(x+2)]=12$

$-3x^2-3x+2x^2+4x=12$

$-x^2+x=12$

$x^2-x+12=0$

Vô nghiệm.

TH2: $-2\le x<-1$

$3x[-(x+1)]-2x(x+2)=12$

$-5x^2-7x=12$

$5x^2+7x+12=0$

Vô nghiệm.

TH3: $x\ge-1$

$3x(x+1)-2x(x+2)=12$

$x^2-x=12$

$x^2-x-12=0$

$(x-4)(x+3)=0$

$x=4$ hoặc $x=-3$.

Chỉ $x=4$ thỏa mãn $x\ge-1$.

Vậy $\boxed{x=4}$.

Xét $2\le x\le3$:

$x-2+3-x+4-x=2$

$5-x=2$

$x=3$.

Xét $3\le x\le4$:

$x-2+x-3+4-x=2$

$x-1=2$

$x=3$.

Vậy $\boxed{x=3}$.

$\left|\frac{11}{5}-x\right|+\left|x-\frac15\right|=\frac{19}{5}$

Vì hai mốc là $\frac15,\frac{11}{5}$:

$\frac{11}{5}-x+\frac15-x=\frac{19}{5}$

$\frac{12}{5}-2x=\frac{19}{5}$

$x=-\frac75$

$\frac{11}{5}-x+x-\frac15=2\ne\frac{19}{5}$

Vô nghiệm.

$x-\frac{11}{5}+x-\frac15=\frac{19}{5}$

$2x-\frac{12}{5}=\frac{19}{5}$

$x=\frac{31}{10}$

Vậy $\boxed{x=-\frac75;\frac{31}{10}}$.

$\left|\frac54x-\frac72\right|=\frac58+\frac35$

$\left|\frac54x-\frac72\right|=\frac{49}{40}$

$\frac54x-\frac72=\pm\frac{49}{40}$

TH1:

$\frac54x-\frac72=\frac{49}{40}$

$\frac54x=\frac{189}{40}$

$x=\frac{189}{50}$

TH2:

$\frac54x-\frac72=-\frac{49}{40}$

$\frac54x=\frac{91}{40}$

$x=\frac{91}{50}$

Vậy $\boxed{x=\frac{91}{50};\frac{189}{50}}$.