23 * [ ( 12 * x + 39 ) : 3 ] = 23 + 40 * 115

23 x [ ( 12 x x + 39 ) : 3 ] = 4623

( 12 x x + 39 ) : 3 = 4623 : 23

( 12 x x + 39 ) : 3 = 201 

12 x x + 39 = 201 x 3 

12 x x + 39 = 603

12 x x = 603 - 39 

12 x x = 564

x = 564 : 12 

x = 47

k mik nha các bn

125 - (23 + 77) + (3x3) - (40+40) = 125 - 100 + 9 - 80

                                                     = 34 - 80

                                                     = -46

Goodluck!!

= 125 - 100 + 9 - 50

= 34 - 80 

= -46

\(125-\left(23+77\right)+\left(3\times3\right)-\left(40+40\right)\)

\(=125-100+9-80\)

\(=25+9-80\)

\(=34-80\)

\(=-46\)

_Chúc bạn học tốt_

`@` `\text {Ans}`

`\downarrow`

`2000 - 40 \times 27 - 40 \times 23`

`= 2000 - 40 \times (27 + 23)`

`= 2000 - 40 \times 50`

`= 40 \times 50 - 40 \times 50`

`=0`

\(2000-40x27-40x23\)

\(=40x50-\left(40x27+40x23\right)\)

\(=40x50-\left[40x\left(27+23\right)\right]\)

\(=40x50-40x50=0\)

$2000-40\times 27-40\times 23$

$=2000-40\times (27+23)$

$=2000-40\times 50$

$=40\times 50-40\times 50$

$=0$

Ta có: 23. [(12.X+39):3] = 4623

⇒(12.X+39):3 = 201

⇒12.X+39 = 603

⇒12.X = 564

⇒X = 47

Lời giải chi tiết:

40 - 23 = 17

40-23=17 bn nhé

=17 dễ quá1111111111111111111111111111111111111111111111111111111111111

\(B=\frac{23^{41}+1}{23^{42}+1}\)

Vì B < 1

\(\Rightarrow B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23(23^{40}+1)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

P/s: Hoq chắc

ta có 

\(B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23\left(23^{40}+1\right)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

\(\Rightarrow B< A\)

Cuối cùng kết luận hộ nhé :v