\(x< y< 0\Rightarrow x,y\)Cùng âm

\(|x|-|y|=100\left(cùngam\right).x< y\Rightarrow x-y=-100\)

9 - 25 = (7 - x) - (25 + 7)

=> -16 = 7-x - 32

=>-25 - x = -16

=> x = -25 + 16

=> x = -9

Có: 9-25=(7-x)-(25+7)

<=> (7-x)-(25+7)=9-25

<=> 7-x-25-7=9-25

<=> -x-25= 9-25

<=> -x-9= -25+25

<=> -x-9=0

<=> -x=9

=> x=-9

Vậy x=-9

Vì \(\left|x+5\right|\ge0\forall x;\left|y-7\right|\ge0\forall y\)

Mà \(\left|x+5\right|+\left|y-7\right|=0\)

\(\Rightarrow\hept{\begin{cases}x+5=0\\y-7=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-5\\y=7\end{cases}}\)

Vậy......

Có: |x+5|+|y-7|=0

=> |x+5|=0 và |y-7|=0

<=>\(\hept{\begin{cases}x+5=0\\y-7=0\end{cases}}\)

<=> x=-5 và y=7

U₁₂(2;3;4;6;12)

Vi 12 chia het cho x nen

Suy ra :x = { 1;2;3;4;6;12}

Vay x ={1;2;3;4;6;12}

\(6⋮x-1\Rightarrow x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}.\)

x - 1 = 1 => x = 2

x - 1 = -1 => x = 0 

x -1 = 2=> x = 3 

x - 1 =....

\(\Rightarrow x\in\left\{2;0;3;4;7\right\}\)

Theo đề bài ta có:

x - 1  Ư(6) = {1;2;3;6}

Mà x  N => x  {2;3;4;7}

Vậy x  {2;3;4;7}

a) 3a + 4b - 5c - 2a - 3b + 5c

= ( 3a - 2a ) + ( 4b - 3b ) - ( 5c - 5c )

= a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= ( 7a - 3a - 4a ) + ( 3b + 2b + b ) - ( 4c + 2c + 2c ) 

= 6b - 8c

a) 3a + 4b - 5c - 2a - 3b + 5c

= (3a - 2a) + (4b - 3b) - (5c - 5c)

= a + b - 0 = a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= (7a - 3a - 4a) + (3b + 2b + b) - ( 4c + 2c + 2c)

= 0 + 6b - 8c = 6b - 8c

Khi chia mot so cho 225 ta duoc so du la 160

=> so do co dang:  \(225k+160\)

Nhan thay:  \(225k⋮85\);   \(160\)khong chia het cho 85

=>  225k + 160 khong chia het cho 85

=> so do khong chia het cho 85