Question

Answer 1

\(IV\\ 1+2+3+...+n=\dfrac{\left(n+1\right)\left(n+1-1\right)}{2}=\dfrac{n\left(n+1\right)}{2}=\dfrac{n^2+n}{2}\)

Lưu ý: \(n+1-1\) là công thức tính số số hạng, còn \(\dfrac{\left(n+1\right)\left(n+1-1\right)}{2}\) là công thức tính tổng dãy số cách đều

Answer 2

\(I\\ 1,\\ a,=x^3+6x^2+12x+8\\ b,=27x^3-54x^2y+36xy^2-8y^3\\ 2,\\ a,=\left(2x-3\right)\left(4x^2+6x+9\right)\\ b,=\left(x+\dfrac{y}{2}\right)\left(x^2-\dfrac{xy}{2}+\dfrac{y^2}{4}\right)\\ II\\ a,\Rightarrow\left(x-4\right)\left(x+4\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ b,\Rightarrow x^3-3x^2+3x-1=0\Rightarrow\left(x-1\right)^3=0\Rightarrow x-1=0\Rightarrow x=1\\ c,\Rightarrow x^3-3x^2+3x-1-3x^2-6x-3-x^3-8=0\\ \Rightarrow-6x^2-3x-12=0\\ \Rightarrow-2x^2-x-4=0\\ \Rightarrow-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{31}{8}=0\\ \Rightarrow-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{31}{8}=0\\ \Rightarrow x\in\varnothing\left[-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{31}{8}\le\dfrac{31}{8}< 0\right]\)

Answer 3

a) \(\left(x+2\right)^3\)

\(=x^3+3.x^2.2+3.x.2^2+2^3\)

\(=x^3+6x^2+12x+8\)

b) \(\left(3x-2y\right)^3\)

\(=\left(3x\right)^3+3.\left(3x\right)^2.2y+3.3x.\left(2y\right)^2+\left(2y\right)^3\)

\(=27x^3+54x^2y+36xy^2+8y^3\)