x = 1

y = 5

\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{7}{6}}{2\dfrac{2}{5}}\)

\(B=\dfrac{\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{7}{6}}{\dfrac{12}{5}}\)

\(B=\dfrac{\dfrac{12}{6}}{\dfrac{12}{5}}\)

\(B=\dfrac{2}{\dfrac{12}{5}}\)

\(B=\dfrac{2\cdot5}{12}\)

\(B=\dfrac{10}{12}=\dfrac{5}{6}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`(x - 5)(3 - x) = 0`

`=>`\(\left[{}\begin{matrix}x-5=0\\3-x=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0+5\\x=3-0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)

Vậy, `x \in` `{3; 5}`

`b)`

`(2x - 8)(5 - x) = 0`

`=>`\(\left[{}\begin{matrix}2x-8=0\\5-x=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=8\\x=5-0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=8\div2\\x=5\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

Vậy, `x \in ` `{4; 5}`

`c)`

`7x(2x - 14) = 0`

`=>`\(\left[{}\begin{matrix}7x=0\\2x-14=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\2x=14\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=14\div2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

Vậy,` x \in`` {0; 7}`

`d)`

`(2x - 4)(6 - 2x) = 0`

`=>`\(\left[{}\begin{matrix}2x-4=0\\6-2x=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=4\\2x=6\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\div2\\x=6\div2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy, `x \in {2; 3}.`

a) (x - 5)(3 - x) = 0

x - 5 = 0 hoặc 3 - x = 0

*) x - 5 = 0

x = 5

*) 3 - x = 0

x = 3

Vậy x = 0; x = 3

b) (2x - 8)(5 - x) = 0

2x - 8 = 0 hoặc 5 - x = 0

*) 2x - 8 = 0

2x = 8

x = 8 : 2

x = 4

*) 5 - x = 0

x = 5

Vậy x = 5; x = 8

c) 7x(2x - 14) = 0

7x = 0 hoặc 2x - 14 = 0

*) 7x = 0

x = 0

*) 2x - 14 = 0

2x = 14

x = 14 : 2

x = 7

Vậy x = 0; x = 7

d) (2x - 4)(6 - 2x) = 0

2x - 4 = 0 hoặc 6 - 2x = 0

*) 2x - 4 = 0

2x = 4

x = 4 : 2

x = 2

*) 6 - 2x = 0

2x = 6

x = 6 : 2

x = 3

Vậy x = 2; x = 3

( x - 5 ) . ( 3 - x ) = 0

⇒  x - 5  = 0  hoặc 3 - x = 0

⇒  x = 5     hoặc    x = 3

Vậy x = 5 hoặc x = 3.

`@` `\text {Ans}`

`\downarrow`

`(x - 5) (3 - x) = 0`

`=>`\(\left[{}\begin{matrix}x-5=0\\3-x=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0+5\\x=3-0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)

Vậy, `x \in {3; 5}.`

25 . { 16 : [ 12 - 8 . 2 ]}

= 25 . { 16 : [ 12 - 16 ]}

= 25 . { 16 : ( - 4 ) }

= 25 .( - 4 )

= -100.

\(x\notin N\)

mà x=n+6; n<5 và nϵN^*

⇒ M không có phần tử nào thỏa điều kiện

a) 37.75+37.45+63.67+63.53

= 37.(75+45)+63.(67+53)

= 37.120+63.120

= 120.(37+63)

= 120.100

= 12000

b) 35.34+35.86+65.75+65.45

= 35.(34+86)+65.(75+45)

= 35.120+65.120

=120.(35+65)

= 120.100

= 12000

c) 78.31+78.24+78.17+22.72

= 78.(31+24+17)+22.72

= 78.72+22.72

= 72.(78+22)

= 72.100

= 7200

$\mathrm{a)\; 37}.75+37.45+63.67+63.53$

$=37.(75+45)+63.(67+53)$

$=37.120+63.120$

$=(37+63).120$

$=100.120$

$=12000$

b) 35⋅34+35⋅86+65⋅75+65⋅45=35(34+86)+65(75+45)=35⋅120+65⋅120=(35+65)⋅120=100⋅120=12000

c) 78 . 31 + 78 . 24 + 78 . 17 + 22 . 72

= 78 . ( 31 + 24 + 17 ) + 22 . 72

= 78 . 72 + 22 . 72

= 72 . ( 78 + 22 )

= 72 .100

= 7200

chúc bạn học tốt ❤❤❤

a) x-5=22 ⇒ x=27 (xϵN)

⇒ Tập hợp có 1 phần tử xϵN

b) 2.y.0=15 ⇒ y.0=15/2 ⇒ y không có phần tử (xϵN)

c) y.0=15 ⇒ y không có phần tử (xϵN)

d) f ϵ {0;5) ⇒ Tập hợp có 2 phần tử fϵN

e) e ϵ {1;2;4;6) ⇒ Tập hợp có 4 phần tử eϵN

A = \(\dfrac{3n+2}{n+1}\) (đk n \(\in\)Z ; n \(\ne\) -1)

A \(\in\) Z  ⇔ 3n + 2 ⋮ n + 1

              3n + 3 - 1 ⋮ n + 1

             3(n +1) - 1 ⋮ n + 1

                           1  ⋮ n + 1

               n + 1 \(\in\) { -1; 1}

               n \(\in\) { -2; 0}

\(A=\dfrac{3n+2}{n+1}=\dfrac{3n+3-2}{n+1}=\dfrac{3\left(n+1\right)-2}{n+1}=3-\dfrac{2}{n+1}\)

Để A có giá trị nguyên ⇒ n+1 là Ư(2)={-1;1;-2;2}

⇒ n+1 ϵ {-1;1;-2;2}

⇒ n ϵ {-2;0;-3;1}

C = \(\dfrac{2018^{2011}+1}{2018^{2019}+1}\)

2018²⁰¹¹ < 2018²⁰¹⁹ ⇒ 2018²⁰¹¹ + 1 < 2018²⁰¹⁹ + 1

⇒ C < 1

D = \(\dfrac{2018^{2017}}{2018^{2013}+1}\) 

Tử số D = 2018²⁰¹⁷ = 2018²⁰¹⁶.( 2017 + 1)

              = 2018²⁰¹⁶.2017 + 2018²⁰¹⁶ > 2018²⁰¹³ + 1

D > 1

Vì C < 1 < D 

Vậy C < D

\(C=2018^{2011}+\dfrac{1}{2018^{2019}+1}\)

\(D=\dfrac{2018^{2017}}{2018^{2013}+1}=\dfrac{2018^{2013}.2018^4}{2018^{2013}+1}=\dfrac{\left(2018^{2013}+1-1\right).2018^4}{2018^{2013}+1}=2018^4-\dfrac{2018^4}{2018^{2013}+1}\)

mà \(2018^4< 2018^{2011}\)

\(\Rightarrow D=2018^4-\dfrac{2018^4}{2018^{2013}+1}< 2018^{2011}-\dfrac{2018^4}{2018^{2013}+1}\)

mà \(2018^{2011}-\dfrac{2018^4}{2018^{2013}+1}< C=2018^{2011}+\dfrac{1}{2018^{2019}+1}\)

\(\Rightarrow D< C\)