Câu hỏi của HEX_trên amazon

Question

Xác định các tập hợp sau bằng cách liệt kê

A = {x   | (2x + 1)(x 2 + x – 1)(2x 2 – 3x + 1) = 0}

B = {x   | 6x 2 – 5x + 1 = 0}

C = {x   | (2x + x 2 )(x 2 + x – 2)(x 2 – x – 1...

Nguyễn Lê Phước Thịnh · Accepted Answer

a: $\left(2x+1\right)\left(x^2+x-1\right)\left(2x^2-3x+1\right)=0$

TH1: 2x+1=0

=>2x=-1

=>$x=-\frac12$

TH2: $x^2+x-1=0$

=>$x^2+x+\frac14-\frac54=0$

=>$\left(x+\frac12\right)^2=\frac54$

=>$\left[\begin{array}{l}x+\frac12=\frac{\sqrt5}{2}\ x+\frac12=-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt5-1}{2}\ x=\frac{-\sqrt5-1}{2}\end{array}\right.$

TH3: $2x^2-3x+1=0$

=>(x-1)(2x-1)=0

=>x=1/2 hoặc x=1

Do đó: A={-1/2;1/2;1;$\frac{\sqrt5-1}{2};\frac{-\sqrt5-1}{2}$ }

b: $6x^2-5x+1=0$

=>$6x^2-2x-3x+1=0$

=>2x(3x-1)-(3x-1)=0

=>(3x-1)(2x-1)=0

=>x=1/3 hoặc x=1/2

=>B={1/3;1/2}

c: $\left(2x+x^2\right)\cdot\left(x^2+x-2\right)\left(x^2-x-12\right)=0$

=>$x\left(x+2\right)\left(x+2\right)\left(x-1\right)\left(x-4\right)\left(x+3\right)=0$

=>$x\left(x+2\right)^2\cdot\left(x-1\right)\left(x-4\right)\left(x+3\right)=0$

=>x∈{0;-2;1;4;-3}

=>C={0;-2;1;4;-3}

g: $x^2-9=0$

=>(x-3)(x+3)=0

=>x∈{3;-3}

=>G={3;-3}

h: $\left(x-1\right)\left(x^2+6x+5\right)=0$

=>(x-1)(x+1)(x+5)=0

=>x∈{1;-1;-5}

=>H={1;-1;-5}

i: $x^2-x+2=0$

=>$x^2-x+\frac14+\frac74=0$

=>$\left(x-\frac12\right)^2+\frac74=0$ (vô lý)

=>x∈∅

=>I=∅

j: $\left(2x-1\right)\left(x^2-5x+6\right)=0$

=>(2x-1)(x-2)(x-3)=0

=>x∈{1/2;2;3}

=>J={1/2;2;3}

k: -3

mà x là số chẵn

nên x∈{-2;0;2;4;6;8;10;12}

=>K={-2;0;2;4;6;8;10;12}

l: $x^2>4$

=>x>2 hoặc x<-2

|x|<10

=>-10

mà x>2 hoặc x<-2

nên x∈{3;4;5;6;7;8;9;-9;-8;-7;-6;-5;-4;-3}

=>L={3;4;5;6;7;8;9;-9;-8;-7;-6;-5;-4;-3}

m: -1

mà k∈Z

nên k∈{0;1;2;3;4}

=>3k∈{0;3;6;9;12}

=>M={0;3;6;9;12}

n: $\begin{cases}x^2-1=0\ x^2-4x+3=0\end{cases}\Rightarrow\begin{cases}x^2=1\ \left(x-1\right)\left(x-3\right)=0\end{cases}$

=>x=1

=>N={1}

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