Bài 1: 

\(101\cdot125+101\cdot25-101\cdot50\)

\(=101\cdot\left(125+25-50\right)\)

\(=101\cdot100\)

\(=10100\)

Bài 2: 

\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)

\(=76\cdot115+24\cdot115\)

\(=115\cdot\left(76+24\right)\)

\(=115\cdot100\)

\(=11500\)

5:

a: =>15x=165

=>x=11

b: =>x(x+1)/2=55

=>x^2+x=110

=>x=10

4: =>7x=56

=>x=8

Bài 1:

101•125+101•25+101•50

= 101•(125+25-50)

=101•100

=10100

Bài 2:

76•115+56•24+59•24

= 76•115+24•(56+59)

= 76•115+24•115

= 115•(76+24)

= 115•100

= 11500

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{99}{101}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{99}{101}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{101}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{99}{101}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{99}{101}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{99}{101}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{99}{202}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{101}\)

\(\Leftrightarrow x=100\)

\(\frac{x}{98}+\frac{x-1}{99}+\frac{x-2}{100}+\frac{1-3}{101}=-4\)

<=>  \(\frac{x}{98}+1+\frac{x-1}{99}+1+\frac{x-2}{100}+1+\frac{x-3}{101}+1=0\)

<=>  \(\frac{x+98}{98}+\frac{x+98}{99}+\frac{x+98}{100}+\frac{x+98}{101}=0\)

<=>  \(\left(x+98\right)\left(\frac{1}{98}+\frac{1}{99}+\frac{1}{100}+\frac{1}{101}\right)=0\)

<=>  \(x+98=0\)  (do 1/98 + 1/99 + 1/100 + 1/101 khác 0)

<=> \(x=-98\)

Vậy...

1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\ \left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\ \left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\ \frac{98}{99}-x=\frac{7}{2}\\ x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)

2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)

1.

Câu 1:

\(A=\frac{\left(1+2+3+...+100\right)x\left(101x102-101x101-51-50\right)}{2+4+6+8+...+2048}\)

\(A=\frac{\left(1+2+3+...+100\right)x\left(101x\left(102-101\right)-\left(50+51\right)\right)}{2+4+6+8+...+2048}\)

\(A=\frac{\left(1+2+3+...+100\right)x\left(101-101\right)}{2+4+6+8+...+2048}\)

\(A=\frac{\left(1+2+3+...+100\right)x0}{2+4+6+8+...+2048}\)

\(A=0\)

       Ta có:Số số hạng từ 2 đến 101 là:

                      (101-2):1+1=100(số hạng)

                 Do đó từ 2 đến 101 có số cặp là:

                       100:2=50(cặp)

\(B=\frac{101+100+99+...+3+2+1}{101-100+99-98+3-2+1}\)

\(B=\frac{5151}{51}\)

\(B=101\)

Câu 2:

a)697:\(\frac{15x+364}{x}\)=17

   \(\frac{15x+364}{x}\)=697:17

    \(\frac{15x+364}{x}\)=41

     15x+364=41x

      41x-15x=364

      26x=364

      x=14

Vậy x=14

b)92.4-27=\(\frac{x+350}{x}+315\)

  \(\frac{x+350}{x}+315\)=341

   \(\frac{x+350}{x}\)=26

    x+350=26

    x=26-350

   x=-324

Vậy x=-324

c, 720 : [ 41 - ( 2x -5)] = 40

    [ 41 - ( 2x -5)] =720:40

     [ 41 - ( 2x -5)] =18

      2x-5=41-18

      2x-5=23

      2x=28

      x=14

Vậy x=14

 d, Số số hạng từ 1 đến 100 là:

       (100-1):1+1=100(số hạng)

Tổng dãy số là:
      (100+1)x100:2=5050

          Mà cứ 1 số hạng lại có 1x suy ra có 100x

Ta có:(x+1) + (x+2) +...+ (x+100) = 5750

         (x+x+...+x)+(1+2+...+100)=5750

          100x+5050=5750

          100x=700

           x=7

Vậy x=7

theo công thức đã cho => 1^3+2^3+..+10^3=(1+2+3+..+10)^2 

=55^2 = (x+1)^2

=> x= 55-1=54

phần b tính ra rồi lm SCP rồi tính ra

Ta có: 33:x(dư 3) => 33-3=30 chia hết cho x

          101: x(dư 11) => 101-11=90 chia hết cho x

=> x thuộc ƯCLN( 30;90) = 30

Vậy x = 30

 Chứng minh rằng : 2^2020 - 2^2016 chia hết cho 15