C

C

C

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{2021}\)

<=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)

<=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)

<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)

<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2042}\)

<=> \(\frac{1}{x+1}=\frac{1}{2021}\)

<=> x + 1 = 2021 

<=> x = 2020

Có phải là bình 6a3 học trường THCS Nguyễn Trãi đúng không 

\(\Leftrightarrow1+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{x\left(x+1\right)}=1+\dfrac{2019}{2021}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2019}{2021}\)

\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2019}{2021}\)

\(\Leftrightarrow\dfrac{2}{x+1}=1-\dfrac{2019}{2021}\)

\(\Leftrightarrow\dfrac{2}{x+1}=\dfrac{2}{2021}\)

\(\Leftrightarrow x+1=2021\)

\(\Leftrightarrow x=2020\)

Đề bạn thiếu 1 số \(x\) nữa đúng không?

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2021}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2021}\)

\(\Rightarrow x+1=2021\)

\(\Rightarrow x=2020\)

Vậy \(x=2020\).

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)

\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4042}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2019}{4042}=\frac{1}{2021}\)

\(\Leftrightarrow x+1=2021\)

\(\Leftrightarrow x=2020\left(tm:x\in N\right)\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+............+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)

\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..........+\frac{1}{x\left(x+1\right)}\right]=\frac{2019}{2021}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+........+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4042}\)

\(\Leftrightarrow\frac{1}{x-1}=\frac{1}{2021}\)

\(\Leftrightarrow x-1=2021\)

\(\Leftrightarrow x=2022\)

Vậy \(x=2022\)

\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{2019}{2021}\)

\(\Leftrightarrow\dfrac{1}{\dfrac{1\cdot2}{2}}+\dfrac{1}{\dfrac{2\cdot3}{2}}+\dfrac{1}{\dfrac{3\cdot4}{2}}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=\dfrac{4040}{2021}\)

\(\Leftrightarrow\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4040}{2021}\)

\(\Leftrightarrow2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{4040}{2021}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2020}{2021}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2020}{2021}\)

\(\Leftrightarrow\dfrac{x}{x+1}=\dfrac{2020}{2021}\)

\(\Leftrightarrow2021x=2020x+2020\Leftrightarrow x=2020\)

Vậy S = {2020}

( 1/2019 +  2011/2020 + 4012/2021) x (1/2 - 1/3-1/6 )

= ( 1/2019 +  2011/2020 + 4012/2021) x 0

=0

Bài 2: 

Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)

\(\Leftrightarrow16x+40=90+30\)

\(\Leftrightarrow16x=80\)

hay x=5

Bài 1 :

[( 35 - 5 ) : 3 ]³ + 3

= [30 : 3]³ + 3

= 10³ + 3

= 1000 + 3

= 1003

Đây nha bạn!!!

Chúc bạn học tốt!!!

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