\(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\left(x\ne-1\right)\\=\dfrac{x^3+2x}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1} \\ =\dfrac{x^3+2x+2x\left(x+1\right)+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{\left(x+1\right)^3}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x^2+2x+1}{x^2-x+1}\)

\(x^2-6x+9=x^2-2.x.3+3^2\\ =\left(x-3\right)^2\)

\(2973780=2^2x3^3x5x5507\)

\(\dfrac{2}{13}< \dfrac{1}{y}< \dfrac{3}{7}\\ \dfrac{6}{39}< \dfrac{6}{6y}< \dfrac{6}{14}\\ 39>6y>14\\ \dfrac{39}{6}>y>\dfrac{14}{6}\)

Mà y là STN nên y nhận giá trị: 3;4;5;6

\(\dfrac{2}{13}< \dfrac{1}{y}< \dfrac{3}{7}\\ \dfrac{6}{39}< \dfrac{6}{y}< \dfrac{6}{14}\\ 39>y>14\)

y là STN nên y nhận giá trị: 15;16;17;18;...;37;38

a) MSC: 35

\(\dfrac{8}{7}=\dfrac{8x5}{7x5}=\dfrac{40}{35}\) và \(\dfrac{16}{35}\)

b) MSC: 36 

\(\dfrac{4}{3}=\dfrac{4x12}{3x12}=\dfrac{48}{36}\) và \(\dfrac{29}{36}\)

c) MSC: 44

\(\dfrac{6}{11}=\dfrac{6x4}{11x4}=\dfrac{24}{44};\dfrac{1}{4}=\dfrac{1x11}{4x11}=\dfrac{11}{44}\)

d) MSC: 60

\(\dfrac{17}{12}=\dfrac{17x5}{12x5}=\dfrac{85}{60};\dfrac{7}{15}=\dfrac{7x4}{15x4}=\dfrac{28}{60}\)

\(y=x^4-2x^2+3,D=ℝ\\ \Rightarrow y'=4x^3-4x=4x\left(x^2-1\right)=4x\left(x-1\right)\left(x+1\right)\\ y'=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Vẽ BBT:

  => D

Tìm n nguyên để 5n+12 chia hết cho n+2

Ta có: 5n+12=5(n+2)+2

Để 5n+12 chia hết cho n+2

Thì: 2 phải chia hết cho n+2 (Vì:5(n+2) luôn chia hết cho n+2 với mọi n nguyên)

=> n+2 thuộc Ư(2)={1;-1;2;-2}

=> n thuộc {-1;-3;0;-4}

\(x:\dfrac{3}{4}=\left(\dfrac{2}{3}+\dfrac{1}{3}\right):\dfrac{3}{4}\\ x:\dfrac{3}{4}=\dfrac{3}{3}:\dfrac{3}{4}\\ x:\dfrac{3}{4}=1:\dfrac{3}{4}\\ x=1\)

\(A=3+3^2+3^3+...+3^{100}\\ \Rightarrow3A=3^2+3^3+3^4+...+3^{101}\\ \Rightarrow3A-A=3^2+3^3+3^4+...+3^{101}-\left(3+3^2+3^3+...+3^{100}\right)\\ \Rightarrow2A=3^{101}-3\\ \Rightarrow2A+3=3^{101}\)

Theo đề: \(2A+3=3^n\)

Do đó n=101