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Bài 4: B
Bài 5:
a: {3;5};{3;7};{5;7};{3;5;7};{3};{5};{7};\(\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\left\{0;1;2;3;4;5\right\}\)
b: \(B=\left\{2;3;4;5\right\}\)
c: \(C=\left\{0;1;-1;2;-2;3;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`#3107.101107`
a,
\(\text{A = }\left\{x\in R\text{ | }\left(2x-x^2\right)\left(3x-2\right)=0\right\}\)
`<=> (2x - x^2)(3x - 2) = 0`
`<=>`\(\left[{}\begin{matrix}2x-x^2=0\\3x-2=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x\left(2-x\right)=0\\3x=2\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2-x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy, `A = {0; 2; 2/3}`
b,
\(\text{B = }\left\{x\in R\text{ | }2x^3-3x^2-5x=0\right\}\)
`<=> 2x^3 - 3x^2 - 5x = 0`
`<=> x(2x^2 - 3x - 5) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-3x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-2x+5x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x^2-2x\right)+\left(5x-5\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x\left(x-1\right)+5\left(x-1\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x+5\right)\left(x-1\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x+5=0\\x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)
Vậy, `B = {-5/2; 0; 1}.`
c,
\(\text{C = }\left\{x\in Z\text{ | }2x^2-75x-77=0\right\}\)
`<=> 2x^2 - 75x - 77 = 0`
`<=> 2x^2 - 2x + 77x - 77 = 0`
`<=> (2x^2 - 2x) + (77x - 77) = 0`
`<=> 2x(x - 1) + 77(x - 1) = 0`
`<=> (2x + 77)(x - 1) = 0`
`<=>`\(\left[{}\begin{matrix}2x+77=0\\x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=-77\\x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=-\dfrac{77}{2}\\x=1\end{matrix}\right.\)
Vậy, `C = {-77/2; 1}`
d,
\(\text{D = }\left\{x\in R\text{ | }\left(x^2-x-2\right)\left(x^2-9\right)=0\right\}\)
`<=> (x^2 - x - 2)(x^2 - 9) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-x-2=0\\x^2-9=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2+x-2x-2=0\\x^2=9\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}\left(x^2+x\right)-\left(2x+2\right)=0\\x^2=\left(\pm3\right)^2\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x\left(x+1\right)-2\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x-2=0\\x+1=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=2\\x=-1\\x=\pm3\end{matrix}\right.\)
Vậy, `D = {-1; -3; 2; 3}.`
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d) \(\sqrt[]{x}>x\)
\(\Leftrightarrow x-\sqrt[]{x}< 0\)
\(\Leftrightarrow\sqrt[]{x}\left(\sqrt[]{x}-1\right)< 0\left(x\ge0\right)\)
\(\Leftrightarrow0< x< 1\)
a) \(P\left(x\right):"x^2-5x+4=0"\)
\(x^2-5x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{1;4\right\}\) để \(P\left(x\right):"x^2-5x+4=0"\) đúng
b) \(P\left(x\right):"x^2-5x+6=0"\)
\(x^2-5x+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{2;3\right\}\) để \(P\left(x\right):"x^2-5x+6=0"\) đúng
c) \(P\left(x\right):"x^2-3x=0"\)
\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}\) để \(P\left(x\right):"x^2-3x=0"\) đúng
d) \(P\left(x\right):"\sqrt[]{x}>x"\)
\(\sqrt[]{x}>x\)
\(\Leftrightarrow x-\sqrt[]{x}< 0\)
\(\Leftrightarrow\sqrt[]{x}\left(\sqrt[]{x}-1\right)< 0\)
\(\Leftrightarrow0< x< 1\)
Vậy \(x\in\left(0;1\right)\) để \(P\left(x\right):"\sqrt[]{x}>x"\) đúng
e) \(P\left(x\right):"2x+3< 7"\)
\(2x+3< 7\)
\(\Leftrightarrow2x< 4\)
\(\Leftrightarrow x< 2\)
Vậy \(x\in(-\infty;2)\) để \(P\left(x\right):"2x+3< 7"\) đúng
f) \(P\left(x\right):"x^2+x+1>0"\)
\(x^2+x+1>0\)
\(\Leftrightarrow x^2+x+\dfrac{1}{4}+\dfrac{3}{4}>0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Leftrightarrow\forall x\in R\) để \(P\left(x\right):"x^2+x+1>0"\) đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
Phương trình \({x^2} - 5x - 6 = 0\) có hai nghiệm là -1 và 6, nên \(A = \{ - 1;6\} \)
Phương trình \({x^2} = 1\) có hai nghiệm là 1 và -1, nên \(B = \{ - 1;1\} \)
Do đó
\(\begin{array}{l}A \cap B = \{ - 1\} ,\\A \cup B = \{ - 1;1;6\} ,\\A\backslash B = \{ 6\} ,\\B\backslash A = \{ 1\} ,\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án: D
(x2 - 4) (x2 - 1) = 0 ⇔ x = ±2; x = ±1 nên A = {-2; -1; 1; 2}
(x2 - 4) (x2 + 1) = 0 ⇔ x2 - 4 = 0 ⇔ x = ±2 nên B = {-2; 2}
x4 - 5x2 + 4)/x = 0 ⇔ x4 - 5x2 + 4 = 0 ⇔ x = ±2; x = ±1 nên D = {-2; -1; 1; 2}
=> A = D
\(D=\left\{1;2;3\right\}\)