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`#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}.`
a: \(A=\left\{0;1;2;3;4;5\right\}\)
b: \(B=\left\{2;3;4;5\right\}\)
c: \(C=\left\{0;1;-1;2;-2;3;-3\right\}\)
Câu 2:
\(\left(A\cup B\right)\cap C=A\cap C=[1;+\infty)\cap\left(0;4\right)=[1;4)\)
Tập này có 3 phần tử nguyên
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
\(a,A=\left\{0;1;2;3;4\right\}\\ b,B=\left\{-16;-13;-10;-7;-4;-1;2;5;8\right\}\\ c,C=\left\{-9;-8;-7;...;7;8;9\right\}\\ d,x^2-3x+1=0\\ \Delta=9-4=5\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{5}}{2}\\x=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\\ \Leftrightarrow D=\left\{\dfrac{3-\sqrt{5}}{2};\dfrac{3+\sqrt{5}}{2}\right\}\)
\(e,2x^3-5x^2+2x=0\\ \Leftrightarrow x\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow E=\left\{0;2\right\}\\ f,F=\left\{0;3;6;9;12;15;18\right\}\)
a/ Đúng, khi \(\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
b/ Sai, ví dụ \(x=0\) thì \(2x^2-3x-5\ne0\)
c/ Sai, khi \(x=-1\)
d/ Sai, \(3x^2+2x-1=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{1}{3}\end{matrix}\right.\) mà \(\left\{-1;-\frac{1}{3}\right\}\notin N\)
e/ Đúng, nhìn câu trên ta thấy pt có 2 nghiệm hữu tỉ
f/ Đúng, vì \(x^2+2x+5=\left(x+1\right)^2+4>0\) \(\forall x\in R\)
thanks