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

\(2^3.19-2^3.14+1^{2020}\\ =2^3.\left(19-14\right)+1\\ =8.5+1\\ =40+1=41\)

\(x^4=x^2\\ \Rightarrow x^4-x^2=0\\ \Rightarrow x^2\left(x^2-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\x^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy các giá trị x thỏa mãn: 0;1;-1

a) \(x^3-\dfrac{3}{2}x^2y+\dfrac{3}{4}xy^2-\dfrac{1}{8}y^3\\ =x^3-3.x^2.\dfrac{1}{2}y+3.x.\left(\dfrac{1}{2}y\right)^2-\left(\dfrac{1}{2}y\right)^3\\ =\left(x-\dfrac{1}{2}y\right)^3\)

b) \(\left(x-y\right)^3+6\left(x-y\right)^2+12\left(x-y\right)+8\\ =\left(x-y\right)^3+3.\left(x-y\right)^2.2+3.\left(x-y\right).2^2+2^3\\ =\left(x-y+2\right)^3\)

Vậy giá trị biểu thức bằng -7 khi y=2x và z=2y

z=2y mà y=2x

=> z=2.2x

=> z=4x

Thay y=2x và z=4x vào biểu thức:

\(\dfrac{x+2x+4x}{x+2x-4x}=\dfrac{7x}{-x}=-7\)