olm chưa death ak ?

a; - \(\dfrac{1}{3}\).(15\(x-9\)) + \(\dfrac{2}{7}\).(- \(x-34\)) = 1 - \(\dfrac{3}{4}\).(-16\(x+4\))

    - 5\(x\) + 3 - \(\dfrac{2}{7}\)\(x\) - \(\dfrac{68}{7}\) = 1 + 12\(x\) - 3 

    12\(x\) + 5\(x\)  + \(\dfrac{2}{7}x\) =  3 - \(\dfrac{68}{7}\)  - 1 + 3 

    17\(x\) + \(\dfrac{2}{7}x\) =  (3 - 1 + 3) - \(\dfrac{68}{7}\)

    \(\dfrac{121}{7}\)\(x\)    =  5 - \(\dfrac{68}{7}\)

     \(\dfrac{121}{7}\) \(x\)  = -  \(\dfrac{33}{7}\)

           \(x\)  =    - \(\dfrac{33}{7}\): \(\dfrac{121}{7}\)

            \(x\) =  - \(\dfrac{3}{11}\)

Vậy  \(x\) = - \(\dfrac{3}{11}\) 

Bg

Ta có: 2x \(⋮\)x - 2  (x \(\inℤ\))

=> 2x - 2(x - 2) \(⋮\)x - 2

=> 2x - (2x - 4) \(⋮\)x - 2

=> 2x - 2x + 4 \(⋮\)x - 2

=> 4 \(⋮\)x - 2

=> x - 2 \(\in\)Ư(4)

Lập bảng: 

Vậy x = {3; 4; 6; 1; 0; -2}

Ta có: \(2x⋮x-2\)

\(\Rightarrow2x-4+4⋮x-2\)

\(\Rightarrow2\left(x-2\right)+4⋮x-2\)

\(\Rightarrow4⋮x-2\)

\(\Rightarrow x-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Rightarrow x\in\left\{3;1;4;0;6;-2\right\}\)

Vậy \(x\in\left\{3;1;4;0;6;-2\right\}\).

3 thứ còn mỗi cái được 101,33 đô

mua ba cái kia hết số tiền là:

509-205=304[đô]

mua mỗi cái hết là:

304:3= 101,3 [đô]

đ/s: 101,3 đô

phao hoặc mỏ neo

mỏ neo ha bn

hello, my name is.....I'm ..... years old.I'm in class... at ......primary schoocl.My birthday is on ........of.......My mom birthday is on..... of....In my free time, I often ........with my mom.I ....... I don't ......

chúc bạn học tốt

mấy câu này dễ mà

ta gọi phần trong ngoặc là A thì ta có

A nhân x = A

     x= A-A

    x=1

Đặt C = \(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\)

=> 100C = \(\frac{100}{1.101}+\frac{100}{2.102}+...+\frac{100}{10.110}\)

=> 100C = \(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\)

=> 100C = \(\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\)

=> C = \(\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)}{100}\)

Lại có B = \(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)

=> 10B = \(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\)

=> 10B = \(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\)

=> 10B = \(\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\)

=> 10B = \(1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\)

=> B = \(\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)}{10}\)

Khi đó \(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right)x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)

<=> C.x = B

<=> \(\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)}{100}x=\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+..+\frac{1}{110}\right)}{10}\)

=> \(x=10\)

Vậy x = 10