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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<x<13
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<x<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<k<5
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}
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<x<13
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<x<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<k<5
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}
a) \(2x^3-3x^2-5x=0\)
\(x\left(x+1\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=-1\left(TM\right)\\x=\dfrac{5}{2}\left(L\right)\end{matrix}\right.\)
\(A=\left\{-1\right\}\)
b) \(x< \left|3\right|\)\(\Leftrightarrow-3< x< 3\)
\(B=\left\{-2;-1;1;2\right\}\)
c) \(C=\left\{-3;3;6;9\right\}\)
a) \(A=\left\{x\in Z|2x^3-3x^2-5x=0\right\}\)
\(2x^3-3x^2-5x=0\)
\(\Leftrightarrow x\left(2x^2-3x-5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow A=\left\{0;-1\right\}\)
b) \(B=\left\{-2;-1;0;1;2\right\}\)
c) \(C=\left\{-3;3;6;9\right\}\)
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: \(D=\varnothing\)
Lời giải:
Đặt $\sqrt{5x^2+10x+1}=a(a\geq 0)$ thì pt trở thành:
$a=7-(x^2+2x)=7-\frac{a^2-1}{5}$
$\Leftrightarrow a=\frac{36-a^2}{5}$
$\Leftrightarrow 5a=36-a^2$
$\Leftrightarrow a^2+5a-36=0$
$\Leftrightarrow (a-4)(a+9)=0$
$\Leftrightarrow a=4$ (do $a\geq 0$)
$\Leftrightarrow 5x^2+10x+1=16$
$\Leftrightarrow 5x^2+10x-15=0$
$\Leftrightarrow 5(x-1)(x+3)=0$
$\Leftrightarrow x=1$ hoặc $x=-3$
Vậy $A=\left\{1;-3\right\}$
Bạn ghi lại đề đi bạn. Với lại cho mình hỏi là đề bài yêu cầu gì vậy?
\(\left(2x+1\right)\left(x^2+x-1\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x^2+x-1=0\\2x^2-3x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\) (pt \(x^2+x-1=0\) ko có nghiệm hữu tỉ nên ko cần quan tâm)
\(A=\left\{-\dfrac{1}{2};\dfrac{1}{2};1\right\}\)
con cãm ơn ạ