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\(A=\left\{25;26;27;28;29;30\right\}\\ B=\left\{11;12;13;14\right\}\\ C=\left\{1;2;5;10\right\}\\ D=\left\{14;16;18;20\right\}\)
\(A=\left\{25;26;27;28;29;30\right\}\)
\(B=\left\{11;12;13;14\right\}\)
\(C=\left\{1;2;5;10\right\}\)
\(D=\left\{14;16;18;20\right\}\)
\(C=\left\{x\in N\text{/}x=m.\left(m+1\right)\right\}\)
Với m = 0 => m.(m + 1) = 0.(0+1) = 0 + 0 = 0
Với m = 1 => m.(m + 1) = 1.(1 + 1) = 1 + 1 = 2
Với m = 2 => m.(m + 1) = 2.(2 + 1) = 4 + 2 = 6
Với m = 3 => m.(m + 1) = 3.(3 + 1) = 9 + 3 = 12
Với m = 4 => m.(m + 1) = 4.(4 + 1) = 16 + 4 = 20
=> Tập hơp C = {0;2;6;12;20}
Với m = 0 thì x = 0 . ( 0 + 1 ) = 0
Với m = 1 thì x = 1 . ( 1 + 1 ) = 2
Với m = 2 thì x = 2 . ( 2 + 1 ) = 6
Với m = 3 thì x = 3 . ( 3 + 1 ) = 12
Với m = 4 thì x = 4 . ( 4 + 1 ) = 20
Kết luận : C = { 0 ; 2 ; 6 ; 12 ; 20 }
a) A = { 13 ; 14 ; 15 }
b) B = { 1 ; 2 ; 3 ; 4 }
c) C = { 13 ; 14 ; 15 }
nha bn
s mik mik tks lại
a/ A = { 13 ; 14 ; 15 }
b/ B = { 1 ; 2 ; 3 ; 4 }
c/ C = { 13 ; 14 ; 15}
`#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}.`
