A  =       \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+ \(\dfrac{1}{3^{99}}\) + \(\dfrac{1}{3^{100}}\)

   3A  = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+ \(\dfrac{1}{3^{99}}\)

3A - A = 1 - \(\dfrac{1}{3^{100}}\)

  2A     =  1 - \(\dfrac{1}{3^{100}}\)

     A     = \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{100}}\) < \(\dfrac{1}{2}\) 

anh ơi bằng -5/7 anh nhé

2x + 3/4 = 1/2 : -5/4 - 1/2 x

2x + 1/2 x = -2/5 - 3/4

5/2 x = -23/20

x = -23/20 : 5/2 = -23/50

Vậy x = -23/50

19/57 . 2 4/7 - 19/57 . 1 5/7 - 19/57

= 19/57 x  ( 2 4/7 - 1 5/7 )

= 19/57 x 6/7

= 2/7

\(\dfrac{19}{57}.2\dfrac{4}{7}-\dfrac{19}{57}.1\dfrac{5}{7}-\dfrac{19}{57}\)

\(\text{=}\dfrac{19}{57}.\left(2\dfrac{4}{7}-1\dfrac{5}{7}-1\right)\)

\(\text{=}\dfrac{19}{57}.\left(\dfrac{-1}{7}\right)\)

\(\text{=}\dfrac{-1}{21}\)

c) \(2021,2345.2020,1234+2021,2345.\left(-2020,1234\right)\)

\(\text{=}2021,2345.\left(2020,1234-2020,1234\right)\)

\(\text{=}2021,2345.0\)

\(\text{=}0\)

d)\(4,75+\left(\dfrac{-1}{2}\right)^3+0,5^2-3.\dfrac{-3}{8}\)

\(\text{=}4,75-\dfrac{1}{8}+\dfrac{1}{4}+\dfrac{9}{8}\)

\(\text{=}\left(4,75+0,25\right)+\left(\dfrac{9}{8}-\dfrac{1}{8}\right)\)

\(\text{=}1+1\)

\(\text{=}2\)

Mình nhầm ở chỗ phần d nha.

Kết quả cuối cùng phải là :

\(\left(4,75+0,25\right)+\left(\dfrac{9}{8}-\dfrac{1}{8}\right)\)

\(\text{=}5+1\)

\(\text{=}6\)

a) \(12,4\cdot6\dfrac{1}{4}+\left(-12,4\right)\cdot\left(-2,5\right)^2\)

\(=12,4\cdot\dfrac{25}{4}-12,4\cdot\dfrac{25}{4}=0\)

b) \(32,125-\left(6,325+12,125\right)-\left(37+13,675\right)=32,125-6,325-12,125-37-13,675\)

\(=\left(32,125-12,125\right)-\left(6,325+13,675\right)-37=20-20-37=-37\)

\(4^x.8^x=1024\)

\(\Rightarrow32^x=1024\)

\(\Rightarrow32^2=1024\)

\(\Rightarrow x=2\)

2\(^x\).2\(^{x+3}\) = 144

2^2\(x\) = 144: 2³

4\(^x\) = 18

nếu \(x\) ≤ 2 ⇒ 4\(^x\) ≤ 4² = 16 < 18 (loại)

Nếu \(x\) ≥ 3 ⇒ 4\(^x\) ≥ 4³ = 64 > 18 (loại)

Vậy \(x\) \(\in\) \(\varnothing\)

\(25^2\cdot2^4=5^{2^2}\cdot2^4=5^4\cdot2 ^4=10^4\)

\(3^8=\left(3^2\right)^4=9^4\)

\(10^4>9^4\Rightarrow25^2\cdot2^4>3^8\)

\(25^2\times2^4\)  và  \(3^8\)

Ta có:

\(25^2\times2^4=625\times16=10000\)

\(3^8=6561\)

Vì: 10000 > 6561

=> \(25^2\times2^4>3^8\)

\(\left(0,25\right)^8=\left[\left(0,5\right)^2\right]^8=\left(0,5\right)^{2.8}=\left(0,5\right)^{16}\)

\(\left(0,125\right)^4=\left[\left(0,5\right)^3\right]^4=\left(0,5\right)^{3.4}=\left(0,5\right)^{12}\)

\(\left(0,0635\right)^2=\left[\left(0,5\right)^3\right]^2=\left(0,5\right)^{3.2}=\left(0,5\right)^6\)