Question

liệt kê các phần tử các tập hợp sau

\(A=\left\{x\in R|2x^3-5x+3=0\right\}\)

\(B=\left\{x\in Q|x=\dfrac{1}{2^a},a\in N,x\ge\dfrac{1}{8}\right\}\)

C là tập hợp các số chính phương k vượt qua 400

Answer 1

A)

\(2x^3-5x+3=0\Leftrightarrow (2x^3-2x)-(3x-3)=0\)

\(\Leftrightarrow 2x(x^2-1)-3(x-1)=0\)

\(\Leftrightarrow 2x(x-1)(x+1)-3(x-1)=0\)

\(\Leftrightarrow (x-1)(2x^2+2x-3)=0\)

\(\Rightarrow \left[\begin{matrix} x=1\\ 2x^2+2x-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{-1\pm \sqrt{7}}{2}\end{matrix}\right.\)

Vậy \(A=\left\{1; \frac{-1+\sqrt{7}}{2}; \frac{-1-\sqrt{7}}{2}\right\}\)

Answer 2

B)

Ta có: \(x=\frac{1}{2^a}\geq \frac{1}{8}\)

\(\Rightarrow 2^a\leq 8\Leftrightarrow 2^a\leq 2^3\)

Mà \(a\in\mathbb{N}\Rightarrow a\in\left\{0;1;2;3\right\}\)

\(\Rightarrow x\in\left\{1; \frac{1}{2}; \frac{1}{4}: \frac{1}{8}\right\}\)

Vậy \(B=\left\{1; \frac{1}{2}; \frac{1}{4}; \frac{1}{8}\right\}\)

C) \(C=\left\{x\in\mathbb{N}|x=a^2,a\in\mathbb{N}, x\leq 400\right\}\)

Ta thấy: \(x=a^2\leq 400\)

\(\Leftrightarrow a^2-400\leq 0\Leftrightarrow (a-20)(a+20)\leq 0\)

\(\Leftrightarrow -20\leq a\leq 20\). Mà \(a\in\mathbb{N}\Rightarrow 0\leq a\leq 20\)

\(\Rightarrow a\in\left\{0;1;2;3;...;20\right\}\)

\(\Rightarrow x\in \left\{0^2;1^2;2^2;3^2;....;20^2\right\}\)

Vậy \(C=\left\{0^2;1^2;2^2;,...; 20^2\right\}\)

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