\(\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2021.2022}\\ =4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2021.2022}\right)\\ =4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\\ =4\left(1-\dfrac{1}{2022}\right)\\ =4.\dfrac{2021}{2022}\\ =\dfrac{4042}{1011}\)

mọi người đừng có hùa theo thế chứ

sai r bạn

\(x^2-4x+4=\left(x-2\right)^2\ge0\)

Mà đề cho \(\left(x-2\right)^2\le0\Rightarrow x-2=0\Rightarrow x=2\)

Ta có: 

\(f\left(x\right)+g\left(x\right)+f\left(x\right)-g\left(x\right)=2x^4+5x^2-3x+x^4-x^2+2x\\ \Rightarrow2f\left(x\right)=3x^4+4x^2-x\\ \Rightarrow f\left(x\right)=\dfrac{3x^4+4x^2-x}{2}\)

\(f\left(x\right)+g\left(x\right)-f\left(x\right)+g\left(x\right)=2x^4+5x^2-3x-x^4+x^2-2x\\ \Rightarrow2g\left(x\right)=x^4+6x^2-5x\\ \Rightarrow g\left(x\right)=\dfrac{x^4+6x^2-5x}{2}\)

a, Ta có:OM=ON\(\Rightarrow\Delta\)OMN cân tại O\(\Rightarrow\widehat{OMN}=\widehat{ONM}\)

Xét ΔOHM và ΔOHN có:

\(\widehat{OHN}=\widehat{OHM}\left(=90^o\right)\)

Chung OH

OM=ON(gt)

\(\Rightarrow\Delta OHM=\Delta OHN\left(ch-gn\right)\)

b, \(\Delta OHM=\Delta OHN\left(cma\right)\Rightarrow HM=HN\) (2 cạnh tương ứng)

\(\dfrac{5}{6}+\dfrac{1}{3}\times2=\dfrac{5}{6}+\dfrac{2}{3}=\dfrac{5}{6}+\dfrac{4}{6}=\dfrac{9}{6}=\dfrac{3}{2}\\ \dfrac{9}{11}:\left(\dfrac{6}{7}-\dfrac{5}{6}\right)=\dfrac{9}{11}:\left(\dfrac{36}{42}-\dfrac{35}{42}\right)=\dfrac{9}{11}:\dfrac{1}{42}=\dfrac{9}{11}\times42=\dfrac{378}{11}\)

\(\dfrac{2x+6}{x-3}=\dfrac{-2}{5}\\ \Leftrightarrow5\left(2x+6\right)=-2\left(x-3\right)\\ \Leftrightarrow10x+30=-2x+6\\ \Leftrightarrow12x+24=0\\ \Leftrightarrow x=-2\)

ĐKXĐ:\(x\ne-1\)

\(\dfrac{1}{x+1}+\dfrac{2x^2+1}{x^3+1}+\dfrac{2x^3-2x^2}{x^2-x+1}=2x\\ \Leftrightarrow\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x^2+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x^2\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2x\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x+1+2x^2+1+2x^2\left(x^2-1\right)-2x\left(x^3+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\)

\(\Rightarrow x^2-x+1+2x^2+1+2x^4-2x^2-2x^4-2x=0\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)