\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

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

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

32-2.(x+1)=42

2.(x+1)=32-42

2(x+1)=-10

x+1=(-10):2

x+1=-5

x=-5-1

x=-6

32-2.(x+1)=42

2.(x+1)=42+32

2.(x+1)=74

x+1=74:2

x+1=37

x=37-1

x=36

chúc bạn hok tốt

\(32-2.\left(x+1\right)=42\)

\(2.\left(x+1\right)=32-42=-10\)

\(x+1=-10:2=-5\)

\(x=-5-1=-6\)

\(42-2\left(32-2^{x+1}\right)=10\)

\(\Leftrightarrow2\left(32-2^{x+1}\right)=42-10=32\)

\(\Leftrightarrow32-2^{x+1}=16\)

\(\Leftrightarrow2^{x+1}=32-16=16\)

\(\Leftrightarrow2^{x+1}=2^4\)

\(\Leftrightarrow x+1=4\)

\(\Rightarrow x=3\)

\(42-2\left(32-2^{x+1}\right)=10\)

\(\Leftrightarrow2^{x+1}=16\)

\(\Leftrightarrow2^{x+1}=2^4\)

=> x = 4

\(42-2.\left(32-2^{x+1}\right)=10\)

\(2.\left(32-2^{x+1}\right)=42-10\)

\(2.\left(32-2^{x+1}\right)=32\)

\(32-2^{x+1}=32:2\)

\(32-2^{x+1}=16\)

\(2^{x+1}=32-16\)

\(2^{x+1}=16\)

\(2^{x+1}=2^4\)

\(x+1=4\)

\(x=4-1\)

\(x=3\)

Làm lại :< 

\(42-2\left(32-2^{x+1}\right)=10\)

\(\Leftrightarrow2^{x+1}=16\)

\(\Leftrightarrow x^{x+1}=2^4\)

\(\Leftrightarrow x=4-1\)

<=> x = 3

\(a,[(2\cdot x-11):3+1]\cdot5=20\\\Rightarrow (2x-11):3+1=20:5\\\Rightarrow (2x-11):3+1=4\\\Rightarrow (2x-11):3=4-1\\\Rightarrow (2x-11):3=3\\\Rightarrow2x-11=3\cdot3\\\Rightarrow2x-11=9\\\Rightarrow2x=9+11\\\Rightarrow2x=20\\\Rightarrow x=20:2=10\)

\(b,(25-2x)^3:5-3^2=4^2\\\Rightarrow(25-2x)^3:5-9=16\\\Rightarrow(25-2x)^3:5=16+9\\\Rightarrow(25-2x)^3:5=25\\\Rightarrow(25-2x)^3=25\cdot5\\\Rightarrow(25-2x)^3=125\\\Rightarrow(25-2x)^3=5^3\\\Rightarrow25-2x=5\\\Rightarrow2x=25-5\\\Rightarrow2x=20\\\Rightarrow x=20:2=10\\Toru\)

\(2,\)

\(a,20-\left[4^2+\left(x-6\right)\right]=90\)

\(\Rightarrow20-16-x+6=90\)

\(\Rightarrow10-x=90\)

\(\Rightarrow x=-80\)

Vậy: \(x=-80\)

\(b,\left(x+3\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;2\right\}\)

\(c,1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\left(x\in N\right)\)

\(\Rightarrow1000:\left(30+2^x-6\right)=25\)

\(\Rightarrow24+2^x=40\)

\(\Rightarrow2^x=16\)

\(\Rightarrow x=4\)

Vậy: \(x=4\)

:))))