a) \(A=\left\{x\in R|x-\sqrt[]{3-2x}=0\right\}\)

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

\(\)\(x-\sqrt[]{3-2x}=0\)

\(\Leftrightarrow\sqrt[]{3-2x}=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\3-2x=x^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+2x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=1\)

\(\Rightarrow A=\left\{1\right\}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

\(\Rightarrow B=\left\{-3;1\right\}\)

Vậy \(A\subset B\)

b) \(A=\left\{x\in N|x^2-2x+1>10\right\}\)

\(B=\left\{x\in N|x>=2\right\}\)

\(x^2-2x+1>10\)

\(\Leftrightarrow\left(x-1\right)^2>\left(\sqrt[]{10}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1< -\sqrt[]{10}\\x-1>\sqrt[]{10}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< 1-\sqrt[]{10}\\x>1+\sqrt[]{10}\end{matrix}\right.\)

\(\Rightarrow A=(-\infty;1-\sqrt[]{10})\cup(1+\sqrt[]{10};+\infty)\)

\(B=[2;+\infty)\)

mà \(1-\sqrt[]{10}< 2< 1+\sqrt[]{10}\)

Vậy 2 tập hợp không có quan hệ gì giữa nhau

a/ Sai vì vs \(x=1\) thì \(x^2=1=x\) (trái vs mệnh đề thứ 2)

b/ Sai vì các nghiệm của PT đều là số âm\(\ne\) N

c/ Đúng vì vs n= 1 thì \(n^2+2n+3=6\) (6 là hợp số)

d/ Sai vì vs x= \(-4< 3\Rightarrow16>9\)

A, Đ

b, S

c, S

d,S

(2x-x^2)(2x^3-3x-2)=0

=>x(2-x)(2x^3-3x-2)=0

=>x=0 hoặc 2-x=0 hoặc 2x^3-3x-2=0

=>\(x\in\left\{0;2;1,48\right\}\)

=>\(A=\left\{0;2;1,48\right\}\)

3<n^2<30

mà \(n\in Z^+\)

nên \(n\in\left\{2;3;4;5\right\}\)

=>B={2;3;4;5}

=>A giao B={2}

=>Chọn B

a) \({x^5}:{x^3} = {x^{5 - 3}} = {x^2}\);

b) \((4{x^3}):{x^2} = (4:1).({x^3}:{x^2}) = 4x\);

c) \((a{x^m}):(b{x^n}) = (a:b).({x^m}:{x^n}) = (a:b).{x^{m - n}}\)(a ≠ 0; b ≠ 0; m, n \(\in\) N, m ≥ n).

a) \({x^2}.{x^4} = {x^{2 + 4}} = {x^6}\).

b) \(3{x^2}.{x^3} = 3.1.{x^{2 + 3}} = 3{x^5}\).

c) \(a{x^m}.b{x^n} = a.b.{x^{m + n}}\) (a ≠ 0; b ≠ 0; m, n \(\in\) N).

Để A là số nguyên dương thì \(\left\{{}\begin{matrix}\sqrt{x}⋮\sqrt{x}-3\\\dfrac{\sqrt{x}}{\sqrt{x}-3}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-3+3⋮\sqrt{x}-3\\\sqrt{x}-3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}-3\in\left\{1;-1;3;-3\right\}\\x>9\end{matrix}\right.\Leftrightarrow x\in\left\{14;36\right\}\)

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

Giải phương trình sau :

 \(\left(x-2x^2\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow x\left(1-2x\right)\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1-2x=0\\x-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\\x=2\end{matrix}\right.\)

\(\Rightarrow A=\left\{0;\dfrac{1}{2};1;2\right\}\)

\(B=\left\{n\in N|3< n\left(n+1\right)< 31\right\}\)

Giải bất phương trình sau :

\(3< n\left(n+1\right)< 31\)

\(\Leftrightarrow\left\{{}\begin{matrix}n\left(n+1\right)>3\\n\left(n+1\right)< 31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n^2+n-3>0\\n^2+n-31< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n< \dfrac{-1-\sqrt[]{13}}{2}\cup n>\dfrac{-1+\sqrt[]{13}}{2}\\\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1-\sqrt[]{13}}{2}\\\dfrac{-1+\sqrt[]{13}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)

Vậy \(B=\left(\dfrac{-1-5\sqrt[]{5}}{2};\dfrac{-1-\sqrt[]{13}}{2}\right)\cup\left(\dfrac{-1+\sqrt[]{13}}{2};\dfrac{-1+5\sqrt[]{5}}{2}\right)\)

\(\Rightarrow A\cap B=\left\{2\right\}\)

a: A={0;1;2;3}

b: B={-16;-13;-10;-7;-4;-1;2;5;8}

c: C={-9;-8;-7;...;7;8;9}

d: \(D=\varnothing\)

Chọn C