a) 2001 x 757 + 2002 x 233

    = 2001 x 757 + (2001 + 1) x 233

    = 2001 x 757 + 2001 x 233 + 1 x 233

    = 2001 x (757+233) + 233

    = 2001 x 1000 + 233

    = 2001000 + 233

    = 2001233

a) 2001x757+2002x243

=2001 x 757+ (2001+1) x233

=2001 x757+2001x233+1 x233

=2001x(757+233)+233

=2001x1000 +233

=201000+233

=2001233

b)(m:1-m x1):(mx2001+m+1)

=(m-m):(mx2001+m+1)

=0:(mx2001+m+1)

=0

   Học tốt

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