Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^4-3x^3-5x^2+12x+4=0\)
\(\Leftrightarrow x^4-2x^3-x^3+2x^2-7x^2+14x-2x+4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-x^2-7x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-3x-1\right)=0\)
mà x là số hữu tỉ
nên x=2 hoặc x=-2
=>A={2;-2}
b: \(x^3+x^2-3x-2=0\)
\(\Leftrightarrow x^3+2x^2-x^2-2x-x-2=0\)
=>(x+2)(x^2-x-1)=0
mà x là số hữu tỉ
nên x=-2
=>B={-2}
c: \(\Leftrightarrow x^4-x^3-x^3+x^2-4x^2+4x-2x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-2x-2\right)=0\)
mà x là số hữu tỉ
nên x=1 hoặc x=-1
=>C={1;-1}
\(A=\left\{-\frac{1}{2};0;2\right\}\)
\(B=\left\{2;3;4;5\right\}\)
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 \(A=\left\{1; \frac{-1+\sqrt{7}}{2}; \frac{-1-\sqrt{7}}{2}\right\}\)
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 \(C=\left\{0^2;1^2;2^2;,...; 20^2\right\}\)
+)
Ta có:
(2x - x2)(2x2 - 3x - 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-x^2=0\\2x^2-3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=2\\x=\frac{-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\frac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\) A = \(\left\{\frac{-1}{2};0;2\right\}\)
Và B = \(\left\{2;3;4;5\right\}\)
Vậy \(A\cap B\) = \(\left\{2\right\}\)
\(x^4-16\left(x^2-1\right)=0\Leftrightarrow x^4-16x^2+16=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=8+4\sqrt{3}\\x^2=8-4\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow A=\left\{-\sqrt{6}-\sqrt{2};\sqrt{2}-\sqrt{6};\sqrt{6}-\sqrt{2};\sqrt{2}+\sqrt{6}\right\}\)
\(2x\le9\Rightarrow x\le\frac{9}{2}\Rightarrow B=\left\{0;1;2;3;4\right\}\)
Bạn coi lại đề, tập hợp A nhìn rất có vấn đề :)
a/ \(\left\{a\right\};\left\{b\right\};\left\{a;b\right\};\varnothing\)
b/ \(\left\{1\right\};\left\{2\right\};\left\{3\right\};\left\{1;2\right\};\left\{1;3\right\};\left\{2;3\right\};\left\{1;2;3\right\};\varnothing\)
c/ \(\left\{0\right\};\left\{1\right\};\left\{2\right\};\left\{3\right\};\left\{0;1\right\};\left\{0;2\right\};\left\{0;3\right\};\left\{1;2\right\};\left\{1;3\right\};\left\{2;3\right\};\left\{0;1;2\right\};\left\{1;2;3\right\};\left\{0;2;3\right\};\left\{0;1;3\right\};\left\{0;1;2;3\right\};\varnothing\)
d/ \(\left\{1\right\};\left\{-2\right\};\left\{1;-2\right\};\varnothing\)
ch ữ đẹp quá :)