\(B=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\=\dfrac{1}{3}\left( \dfrac{3}{3.6}+\dfrac{3}{6.9}+\dfrac{3}{9.12}+...+\dfrac{3}{30.33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}.\dfrac{10}{33}\\ =\dfrac{10}{99}\)

\(b,y^2+4^x+2y-2^{x+1}+2=0 \\ \Leftrightarrow\left(y^2+2y+1\right)+\left(4^x-2^{x+1}+1\right)=0\\ \Leftrightarrow\left(y+1\right)^2+\left[\left(2^x\right)^2-2^x.2.1+1^2\right]=0\\ \Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\)

Ta có:
`(y+1)^2 >= 0 forall y`

`(2^x -1)^2 >= 0 forall x`

`=> (y+1)^2 + (2^x -1)^2 >= 0 forall x,y`

Dấu "=" xảy ra `<=>{(y+1=0),(2^x -1=0):}`

`<=>{(y=-1),(2^x =1):}`

`<=>{(y=-1),(x=0):}`

Vậy `(x,y)=(0,-1)`

\(\left(x^2-x\right)\left(x^2+3x+2\right)=24\\ \Leftrightarrow x\left(x-1\right)\left(x^2+x+2x+2\right)=24\\ \Leftrightarrow x\left(x-1\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]=24\\ \Leftrightarrow x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\\ \Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\\ \Leftrightarrow\left(x^2+x\right)\left(x^2-x+2x-2\right)-24=0\\ \Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\\ \Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\\ \Leftrightarrow\left(x^2+x\right)^2-6\left(x^2+x\right)+4\left(x^2+x\right)-24=0\\ \Leftrightarrow\left(x^2+x\right)\left(x^2+x-6\right)+4\left(x^2+x-6\right)=0\)

\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3x-2x-6=0\\x^2+x+\dfrac{1}{4}+\dfrac{15}{4}\end{matrix}\right.=0\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x+3\right)-2\left(x+3\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy pt có tập nghiệm `S={2;-3}`

\(a,ĐKXĐ:9-x^2\ge0\\ \Leftrightarrow x^2\le9\\ \Leftrightarrow-3\le x\le3\\ b,ĐKXĐ:\dfrac{1}{x^2-4}\ge0\\ \Leftrightarrow x^2-4\le0\\ \Leftrightarrow x^2\le4\\ \Leftrightarrow-2\le x\le2\\ c,ĐKXĐ:\left\{{}\begin{matrix}\sqrt{x}+2\ne0\left(luôn.đúng\right)\\x\ge0\end{matrix}\right.\\ \Leftrightarrow x\ge0\\ d,ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

\(a,\left(\dfrac{1}{1011}\right)^{x+1}.2022^{x+1}=32\\ \Leftrightarrow\left(\dfrac{1}{1011}.2022\right)^{x+1}=2^5\\ \Leftrightarrow2^{x+1}=2^5\\ \Leftrightarrow x+1=5\\ \Leftrightarrow x=4\)

\(b,\left(2x+\dfrac{1}{5}\right)^3=\left|0,3-\dfrac{1}{2}\right|^5:\left(2x+\dfrac{1}{5}\right)^2\\\Leftrightarrow\left(2x+\dfrac{1}{5}\right)^3.\left(2x+\dfrac{1}{5}\right)^2=\left|0,3-0,5\right|^5\\ \Leftrightarrow \left(2x+\dfrac{1}{5}\right)^5=\left|-0,2\right|^5\\ \Leftrightarrow\left(2x+\dfrac{1}{5}\right)^5=0,2^5\\ \Leftrightarrow2x+0,2=0,2\\ \Leftrightarrow2x=0\\ \Leftrightarrow x=0\)

`a, 24:x +16:x=5`

`(24+16):x=5`

`40:x=5`

`x=40:5`

`x=8`

`b,72:x-27:x=5`

`(72-27):x=5`

`45:x=5`

`x=45:5`

`x=9`

\(C=\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}-\dfrac{1}{49.99}\\ =2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\right)-\dfrac{1}{49.99}\\ =2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{49.99}\\ =2.\left(1-\dfrac{1}{99}\right)-\dfrac{1}{49.99}\\ =\dfrac{2.98}{99}-\dfrac{1}{49.99}\\ =\dfrac{196}{99}-\dfrac{1}{4851}\\ =\dfrac{97}{49}\)