⇔ 1999 + \(\overline{19a8}\) ⋮ 1997

⇔ 1999 + 1908 + 10a ⋮ 1997

⇔ 1997 + 2+ 1997 - 89 + 10a ⋮ 1997

⇔            10a - 87 ⋮ 1997

vì là chữ số thuộc hàng chục nên 0 ≤ a ≤ 9

⇔ -87 ≤ 10a - 87 ≤ 10x 9 - 87 = 3

vậy 10a - 87 không chia hết cho 1997

⇔ 1999 + \(\overline{19a8}\) không chia hết cho 1997 với mọi a ϵ {0;1;2;3;4;5;6;7;8;9}

hay không có giá trị nào của a thỏa mãn đề bài

so sánh hã bạn?

hello

`x - ( 5/4 - 7/5) = 9/20`

`x- (-3)/20 = 9/20`

`x + 3/20  = 9/30`

`x = 9/30 - 3/20`

`x=6/20`

`x= 3/10`

__________________________

`9-x = 8/7 - (-7/8)`

`9 - x = 8/7 + 7/8`

`9 - x= 113/56`

`x = 9 - 113/56`

`x= 391/56`

= x - ( -3/20) = 9/20

= x= 9/20 + ( -3/20)

= x=3/10

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= 9-x=113/56

= x=9- 113/56

= x=391/56

`-1,2+(-0,8)+0,25+5,75-2021`

`=[-1,2+(-0,8)]+(0,25+5,75)-2021`

`=-2+6-2021`

`=4-2021=-2017`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`17/11-(6/5-16/11)+26/5`

`=17/11-6/5+16/11+26/5`

`=(17/11+16/11)+(-6/5+26/5)`

`=33/11+20/5`

`=3+4=9`

` -1,2 + (-0,8) + 0,25 + 5,75 - 2021`

` =  (-1,2 + (-0,8) )+ (0,25 + 5,75) - 2021`

`= -2 + 6 - 2021`

` = 4 - 2021`

` = -2017`

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`17/11 - ( 6/5 - 16/11) + 26/5`

`= 17/11 - 6/5 + 16/11 + 26/5`

` = ( 17/11 + 16/11) + ( -6/5 + 26/5)`

` = 33/11 + 20/5`

`= 3 + 4`

` =7`

2⁴ x 8³ = 2⁴x( 2³)³ = 2¹³

`2^4 .8^3`

`=2^4 .(2^3)^3`

`=2^4 .2^9`

`=2^[4+9]=2^13`

\(\left(x-2\right)^2=1\\ \left(x-2\right)^2=1^2\\ x-2=1\\ x=3\)

\(\left(x-2\right)^2=1\)

\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

 a, ( x - \(\dfrac{1}{2}\))⁰ = 0 ⇔ x - 1/2 = 0 ⇒ x =1/2

b, (x - 2)² = 1 ⇒ x - 2 = 1 hoặc x -2 = -1

x -2 = 1 ⇒ x = 2 + 1 = 3

x - 2 = -1 ⇒ x = 2 - 1 = 1

vậy x ϵ {1; 3}

c, (2x - 1 )³ = - 8 ⇒ (2x - 1)³ = (-2)³ ⇒ 2x - 1 = -2

⇒ 2x = 1 - 2 = -1⇒ x = -1/2

d, (x + \(\dfrac{1}{2}\))² = \(\dfrac{1}{16}\) ⇒ ( x+ \(\dfrac{1}{2}\))²  =( \(\dfrac{1}{4}\))²

                                x + \(\dfrac{1}{2}\) = \(\dfrac{1}{4}\) hoặc x + \(\dfrac{1}{2}\) = - \(\dfrac{1}{4}\)

x + \(\dfrac{1}{2}\) = \(\dfrac{1}{4}\)  ⇒ x = \(\dfrac{1}{4}\)-  \(\dfrac{1}{2}\)= \(\dfrac{-1}{4}\)

x + \(\dfrac{1}{2}\) = - \(\dfrac{1}{4}\) ⇒  x = -\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) = \(\dfrac{-3}{4}\)

vậy x ϵ {-\(\dfrac{3}{4}\), - \(\dfrac{1}{4}\)}

a. \(\left(5x-1\right)^6=3^6\\ 5x-1=3\\ 5x=4\\ x=\dfrac{4}{5}\)

b. \(\left(2x+1\right)^3=\left(-0,1\right)^3\\ 2x+1=-0,1\\ 2x=-0,1-1\\ 2x=-1,1\\ x=-0,55\)

(5x-1)⁶ = 729

(5x-1)⁶ = 3⁶

5x - 1 = - 3 hoặc 5x - 1 = 3

5x - 1 = -3

x = (-3 + 1) : 5 = -2/5

5x - 1 = 3 

5x = 3 + 1

5x = 4 

x = 4/5

vậy x ϵ {-2/5; 4/5}

b, (2x + 1)³ = -0,001 = (-0,1)³

2x + 1 = -0,1

2x + 1 = -0,1 - 1 = -1,1

x = -1,1 : 2 

x = -0.55

x < \(\dfrac{x+y}{2}\) < y 

ta có x + x < x + y < y + y (vì x< y)

⇔2x < x + y < 2y

⇔\(\dfrac{2x}{2}\) < \(\dfrac{x+y}{2}\) < \(\dfrac{2y}{2}\)

⇔ x < \(\dfrac{x+y}{2}\) < y (ddpcm)

Có x<y

=>2x<x+y

=>x<x+y/2

Có x<y

=>x+y<2y

=>x+y/2<y

Vậy x,y là số hữu tỉ và x<y thì x<x+y/2<y