Đặt A=1+2+2²+2³+...+2¹⁰⁰

suy ra 2A=2+2²+2³+...+2¹⁰⁰

suy ra 2A-A=(2+2²+2³+...+2¹⁰¹)-(1+2+2²+2³+...+2¹⁰⁰⁾

                 =2¹⁰¹-1

Vậy 1+2+2²+2³+...+2¹⁰⁰=2¹⁰¹-1

 A=1+2+2^2+2^3+...+2^100

2A=2+2²+2³+2⁴+...+2¹⁰¹

2A-A=2¹⁰¹-1

Vậy A= 2¹⁰¹-1

(x+1)+(x+2)+...+(x+98)+(x+99)=9900

x.99+(1+2+3+...+98+99)=9900

x.99+[(99-1):1+1].(99+1):2=9900

x.99+99.100:2

x.99+99.50=9900

x.99+4950=9900

x.99=9900-4950

x.99=4950

x=4950:99

x=50

chúc bạn học tốt nha

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+98\right)+\left(x+99\right)=9900\)

\(\left(x+x+x+...+x+x\right)+\left(1+2+3+...+99\right)=9900\)

  \(\left(99\cdot x\right)+\left(100\times99\div2\right)=9900\)

\(99x+4950=9900\)

\(99x=9900-4950\)

\(x=4950\div99\)

\(x=50\)

ai do k minh nha

Ta có :\(\frac{x}{6}-\frac{7}{y}=\frac{1}{12}\)(y khác 0)

=> \(\frac{xy-42}{6y}=\frac{1}{12}\)

=> 12(xy - 42) = 6y

=> 12xy - 504 = 6y

=> 12xy - 6y = 504

=> 2xy - y = 84

=> y(2x - 1) = 84

Ta có 84 = 1.84 = (-1).(-84) = 42.2 = (-42).(-2) = 21.4 = (-21).(-4) = 7.12 = (-7).(-12) = (-3).(-28) = 28.3 = 14.6 = (-14).(-6)

Lập bảng xét 24 trường hợp

Vậy các cặp (y;x) thỏa mãn là : (84;1) ; (4 ; 11) ; (12 ; 4) ; (28 ; 2) ; (-4 ; - 10) ; (-12 ; -3) ; (-28 ; -1) ; (-84 ; 0)

`x-(5/6 -x) =x-2/3`

`x-5/6 +x -x+2/3 =0`

`x = 5/6-2/3 = 5/6 -4/6 = 1/6`

bạn có thể giải chi tiết hơn dc ko TV Cuber

\(\Rightarrow99x+\left(1+2+3+...+98+99\right)=9900\)(vì có 99 số hạng nha)

\(\Rightarrow99x+4950=9900\)

\(\Rightarrow99x=4950\)

\(\Rightarrow x=50\)

\(x+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+\left(x+\frac{1}{12}\right)+...+\left(x+\frac{1}{9900}\right)=2\)

=> \(x+\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+\left(x+\frac{1}{3.4}\right)+...+\left(x+\frac{1}{99.100}\right)=2\)

 => \(\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2\)(100 hạng tử x)

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

=>  \(100x+1-\frac{1}{100}=2\)

=> \(100x+\frac{99}{100}=2\)

=> \(100x=\frac{101}{100}\)

=> \(x=\frac{101}{10000}\)

\(\dfrac{x-6}{50}+\dfrac{x-6}{51}=\dfrac{x-6}{52}+\dfrac{x-6}{53}\)

\(\Rightarrow\dfrac{x-6}{51}+\dfrac{x-6}{50}-\dfrac{x-6}{52}-\dfrac{x-6}{53}=0\)

\(\Rightarrow\left(x-6\right)\left(\dfrac{1}{50}+\dfrac{1}{51}-\dfrac{1}{52}-\dfrac{1}{53}\right)=0\)

\(\Rightarrow x-6=0\) \(\Rightarrow x=6\)

  Vậy ...