\(\overrightarrow{BM}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{AC}\)

\(y=\dfrac{cotx}{cosx-1}\)

Đk:\(cosx-1\ne0\Leftrightarrow cosx\ne1\)\(\Leftrightarrow x\ne k\pi,k\in Z\)

\(D=R\backslash\left\{k\pi;k\in Z\right\}\)

Ý C

\(sinx=-\dfrac{4}{3}< -1\)

\(\Rightarrow\)Phương trình vô nghiệm

Do ABCD là hình bình hành

\(\Rightarrow\overrightarrow{AB}=\overrightarrow{DC}\)

\(\overrightarrow{BC}-\overrightarrow{AB}=\overrightarrow{BC}-\overrightarrow{DC}=\overrightarrow{BC}+\overrightarrow{CD}=\overrightarrow{BD}\)

a)\(S=\left\{2;3;5;7;11;13;17;19\right\}\)

b)\(A=\left\{\dfrac{1}{3};3\right\}\)

c)\(B=\left\{0;1;2;3\right\}\)

d)\(C=\left\{1\right\}\)

e)\(D=\left\{\varnothing\right\}\)

f)\(D=\left\{-\dfrac{1}{2};\dfrac{1}{2}\right\}\)

Đề sai,biểu thức trong căn <0 

\(=\left(x^2+5x+8\right)\left(x^2+4x+2x+8\right)=\left(x^2+5x+8\right)\left[x\left(x+4\right)+2\left(x+4\right)\right]\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\) 

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8\right)^2+2x\left(x^2+4x+8\right)+x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)\left(x^2+4x+8+2x\right)+x\left(x^2+4x+8+2x\right)\)

\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)

\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)