\(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a+b\right)\left(\left(b+c\right)\left(c-a\right)+\left(c+a\right)\left(b-c\right)\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a+b\right)\left(2bc-2ac\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)} =\dfrac{-2c\left(a+b\right)\left(a-b\right)+\left(a-b\right)\left(b+c\right)\left(c+a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(-2ac-2bc+bc+ab+c^2+ac\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(ab-bc-ac+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)

\(C=\dfrac{9x^2-16}{3x^2-4x}\)
ĐKXĐ: 3x²-4x\(\ne\)0 \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{4}{3}\end{matrix}\right.\)
C=0 <=> x=4/3(ktm đkxđ) hoặc x=-4/3(tm đkxđ) =>x=-4/3
\(C=\dfrac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\dfrac{3x+4}{x}=3+\dfrac{4}{x}\)
Thôi câu khác tương tự nhá

\(B=\dfrac{x^2-9}{x^2-6x+9}=\dfrac{x^2-9}{\left(x-3\right)^2}\)
ĐKXĐ: (x-3)² \(\ne\)0 <=> \(x\ne3\)
B=0 <=> x²-9=0 <=> x=3(ko tm đkxđ) or x=-3(tm đkxđ)=>x=-3
\(B=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)

|5x - 2|< hoặc = 13 <=> 5x - 2 <= 13 hoặc 5x - 2 <= -13 \(\Rightarrow\)x <= 3 hoặc x = -11/5

 |7x + 3| = 66 <=> 7x + 3 =66 hoặc 7x + 3 =-66 \(\Rightarrow\)x = 9 hoặc x = - 69/7

Đề sai sai kiểu j ấy, ko có quy luật j cả

Nhìu thế này khó mà làm hết