1A;2B;3B;4B;5B;6D

\(A\left(4;2\right),B\left(4;-5\right);C\left(1;-1\right)\) \(\Rightarrow\overrightarrow{AB}=\left(0;-7\right);\overrightarrow{AC}=\left(-3;-3\right)\) 

 \(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}\) \(=0.\left(-3\right)+\left(-7\right).\left(-3\right)=21\)

C/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\) (*)

Thật vậy , (*) \(\Leftrightarrow\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(a+2\right)\left(c+2\right)=\left(a+2\right)\left(b+2\right)\left(c+2\right)\)

\(\Leftrightarrow ab+bc+ac+4\left(a+b+c\right)+12=abc+2\left(ab+bc+ac\right)+4\left(a+b+c\right)+8\)

\(\Leftrightarrow ab+bc+ac+abc=4\) (Đ)

=> (*) đúng ( đpcm ) 

Câu 34 . G là trọng tâm \(\Delta A'B'C'\)     . Suy ra :  

\(\overrightarrow{AG}=\overrightarrow{AA'}+\overrightarrow{A'G}=\overrightarrow{AA'}+\dfrac{1}{3}\left(\overrightarrow{A'B'}+\overrightarrow{A'C'}\right)=\dfrac{1}{3}\left(\overrightarrow{3AA'}+\overrightarrow{AB}+\overrightarrow{AC}\right)\) 

Chọn A

Câu 35 . Chọn C 

Do \(x\rightarrow-\infty\Rightarrow\left|x\right|=-x\) 

a. \(lim_{x\rightarrow-\infty}\dfrac{\left|2x\right|^3-\left|x\right|+1}{4\left|x^3\right|+x^2+1}=lim_{x\rightarrow-\infty}\dfrac{-8x^3+x+1}{-4x^3+x^2+1}\)\(=lim_{x\rightarrow-\infty}\dfrac{-8+\dfrac{1}{x^2}+\dfrac{1}{x^3}}{-4+\dfrac{1}{x}+\dfrac{1}{x^3}}=2\)

b. \(lim_{x\rightarrow-\infty}\dfrac{\left|x\sqrt{x^2+3}+1\right|}{\left|x^2-1\right|+x}=lim_{x\rightarrow-\infty}\dfrac{\left|\dfrac{\sqrt{x^2+3}}{x}+\dfrac{1}{x^2}\right|}{\left|1-\dfrac{1}{x^2}\right|+\dfrac{1}{x}}\)  \(=lim_{x\rightarrow-\infty}\dfrac{\left|\dfrac{1}{x^2}-\sqrt{1+\dfrac{3}{x^2}}\right|}{\left|1-\dfrac{1}{x^2}\right|+\dfrac{1}{x}}=\dfrac{-1}{1}=-1\)

1 ) \(lim_{x\rightarrow+\infty}\dfrac{3x^2+5}{x^3-x+2}=lim_{x\rightarrow+\infty}\dfrac{\dfrac{3}{x}+\dfrac{5}{x^3}}{1-\dfrac{1}{x^2}+\dfrac{2}{x^3}}=0\)

2 ) \(lim_{x\rightarrow-\infty}\dfrac{2x^2\left(3x^2-5\right)^3\left(1-x\right)^5}{3x^{14}+x^2-1}\)  \(=lim_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}\left(3-\dfrac{5}{x^2}\right)^3\left(\dfrac{1}{x}-1\right)^5}{3+\dfrac{1}{x^{12}}-\dfrac{1}{x^{14}}}=0\)

3 ) \(lim_{x\rightarrow+\infty}\dfrac{3x-\sqrt{2x^2+5}}{x^2-4}=lim_{x\rightarrow+\infty}\dfrac{\left(7x^2-5\right)}{\left(3x+\sqrt{2x^2+5}\right)\left(x^2-4\right)}\)

\(=lim_{x\rightarrow+\infty}\dfrac{\dfrac{7}{x}-\dfrac{5}{x^3}}{\left(3+\sqrt{2+\dfrac{5}{x^2}}\right)\left(1-\dfrac{4}{x^2}\right)}=0\)

Ta có : \(A=lim_{x\rightarrow1}\dfrac{\sqrt{2x-1}-x}{x^2-1}=lim_{x\rightarrow1}\dfrac{2x-1-x^2}{\left(x^2-1\right)\left(\sqrt{2x-1}+x\right)}\)  \(=lim_{x\rightarrow1}\dfrac{-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{2x-1}+x\right)}\)  \(=lim_{x\rightarrow1}\dfrac{1-x}{\left(x+1\right)\left(\sqrt{2x-1}+x\right)}=0\)