\(\dfrac{2^7 . 9^3}{6^5 . 8^2}\)

\(=\dfrac{2^7 . (3^2)^3}{2^5 . 3^5 . (2^3)^2}\)

\(=\dfrac{2^7 . 3^6}{2^11 . 3^5}\)

\(=\dfrac{3}{2^4}=\dfrac{3}{16}\)

A = {x = 50k| kϵZ}

\(Ư_{\left(50\right)}=\left\{0;50;100;150;200;250;300;350;...\right\}\)

a, X=0;1;2;3

b, X= 7;8;9;10

c, X= 12;14;16;18

d, Tất cả các số tự nhiên khác 0

`a)x < 4` mà \(x \in N\)

  \(=>x \in {0;1;2;3}\)

\(b)7 \le x \le 10\) mà \(x \in N\)

  \(=>x \in {7;8;9;10}\)

\(c)12 \le x < 20\) mà \(x \in N\) và `x` là số tự nhiên

    \(=>x \in {12;14;16;18}\)

\(d) x \notin N\)* mà \(x \in N\)

   $=>x=0$

\(-6-10x-8x+60=36\Leftrightarrow-18x=-18\Leftrightarrow x=1\)

`2(-3-5x)-4(2x-15)=36`

`-6-10x-8x+60=36`

`-10x-8x=36+6-60`

`-18x=-18`

`x=-18:(-18)=1`

`A = { 22,23,24,25,26}`

`B = {2,3,4,5,6}`

`C = {1}`

`D = {2,3}`

A = { 22;23;24;25;26}

B = { 2;3;4;5;6}

C = { 1 }

D = {1;2;3;4}

a,

Theo đề ra, ta có: \(\widehat{xOt}=50^o,\widehat{xOy}=100^o\Rightarrow\widehat{xOt}< \widehat{xOy}\)

Vậy tia Ot nằm giữa tia Ox và Oy (*).

b,

Từ (*) \(\Rightarrow\widehat{xOt}+\widehat{tOy}=\widehat{xOy}\)

\(\Rightarrow50^o+\widehat{tOy}=100^o\)

\(\Rightarrow\widehat{tOy}=\widehat{xOt}=50^o\) (**).

c,

Từ (*)(**) => Tia Oy là tia phân giác của góc xOy.

P = {1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11}

P = { x ϵ N^*| x<12}

\(P=\left\{1;2;3;4;5;6;7;8;9;11\right\}\)

\(P=(x\inℕ^∗\)/ \(x< 12)\)

`4/5 + 5/7 . x=1/6`

`5/7.x = 1/6 - 4/5`

`5/7 . x= (-19)/30`

`x=(-19)/30 : 5/7`

`x=(-19)/30 xx 7/5`

`x=(-133)/150`

\(\dfrac{5}{7}x=\dfrac{1}{6}-\dfrac{4}{5}=-\dfrac{19}{30}\Leftrightarrow x=-\dfrac{19}{30}:\dfrac{5}{7}=-\dfrac{133}{150}\)

A = (1+3+5+7+....+2017)(135135x 137 -137137x135)

A =(1+3+5+7+.....2017) ( 135 x137 x 1001 - 135 x 137 x 1001)

A = (1+3+5+7+....+2017). 0

A = 0

`(1+3+5+7+.....+2017).(135135 xx 137 - 137137 xx 135)`

`=(1+3+5+7+.....+2017).(18513495-18513495)`

`=(1+3+5+7+.....+2017).0`

`=0`

\(6x\left(x-4\right)=\dfrac{14}{3}\Leftrightarrow6x^2-24x-\dfrac{14}{3}=0\Leftrightarrow x=\dfrac{6\pm\sqrt{43}}{3}\)