A= x= 3;4

B x=1;2

\(a,\left(x-3\right)^{27}=\left(x-3\right)^{127}\)

\(\Leftrightarrow\left(x-3\right)^{127}-\left(x-3\right)^{27}=0\)

\(\Leftrightarrow\left(x-3\right)^{27}\left[\left(x-3\right)^{100}-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^{27}=0\\\left(x-3\right)^{100}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{100}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x-3=\pm1\end{cases}\Rightarrow}x\in\left\{2;3;4\right\}}\)

Vậy \(x\in\left\{2;3;4\right\}\)

\(b,\left(x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy \(x\in\left\{1;2\right\}\)

a, \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)

\(\Rightarrow3\left(x-1-4x^2+4x\right)+4\left(3x^2+9x-2x-6\right)=-27\)

\(\Rightarrow15x-3-12x^2+12x^2+28x-24=-27\)

\(\Rightarrow43x=-27+24+3\Rightarrow x=0\)

Các câu còn lại làm tương tự! Phá tan tành hoa loa kèn nhà nó ra!

Chúc bạn học tốt!!!

a. \(lim_{x\rightarrow3}\dfrac{x^3-27}{3x^2-5x-2}=\dfrac{3^3-27}{3.3^2-5.3-2}=\dfrac{0}{10}=0\)

b. \(lim_{x\rightarrow2}\dfrac{\sqrt{x+2}-2}{4x^2-3x-2}=\dfrac{\sqrt{2+2}-2}{4.2^2-3.2-2}=\dfrac{0}{8}=0\)

c. \(lim_{x\rightarrow1}\dfrac{1-x^2}{x^2-5x+4}=lim_{x\rightarrow1}\dfrac{\left(1-x\right)\left(x+1\right)}{\left(x-1\right)\left(x-4\right)}=lim_{x\rightarrow1}\dfrac{-\left(x+1\right)}{x-4}=\dfrac{-\left(1+1\right)}{1-4}=\dfrac{2}{3}\)

d. Câu này mình chịu, nhìn đề hơi lạ so với bình thường hehe