\(\frac{1}{1}=1\)

Viết lại: 1+2+1+2+3+...+1+2+...+10

=2+6+...+55

=163

Hội con 🐄 chúc bạn học tốt!!!

\(D=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+10}\)

\(D=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{\left[1+10\right]\cdot10}{2}}\)

\(D=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{55}\)

\(D=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}\)

\(\frac{D}{2}=2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\right]\)

\(D=2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{10\cdot11}\right]\)

\(D=2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right]\)

\(D=2\left[\frac{1}{2}-\frac{1}{11}\right]\)

\(D=\frac{9}{11}\)

Vậy D = 9/11

Bỏ chỗ \(\frac{D}{2}\)nhé Hoa Thiên Cốt

mk ko bt

2^2.3^1-(1^2021+2021^2) : (-2)

= 4.3-(1+4084441 ) : (-2)

= 12-4084442:(-2)

= 12-(-2042221)

= 20421133

\(n=1\) không thỏa mãn.

ab

Lời giải:
$x-\frac{x}{3}\times \frac{3}{2}=2-\frac{1}{2}$

$x-x\times \frac{1}{2}=\frac{3}{2}$

$x\times (1-\frac{1}{2})=\frac{3}{2}$

$x\times \frac{1}{2}=\frac{3}{2}$

$x=\frac{3}{2}: \frac{1}{2}=3$

       \(\frac{2}{5}\times\frac{1}{2}-\frac{2}{5}\times\frac{1}{3}-\frac{2}{5}\times\)\(\frac{1}{6}\)

\(=\frac{2}{5}\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(=\frac{2}{5}\times\frac{0}{6}\)

\(=\frac{2}{5}\times0\)

\(=0\)

\(\frac{2}{5}\times\frac{1}{2}-\frac{2}{5}\times\frac{1}{3}-\frac{2}{5}\times\frac{1}{6}\)

\(=\frac{2}{5}\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(=\frac{2}{5}\times0\)

\(=0\)

\(\frac{2}{5}\times\frac{1}{2}-\frac{2}{5}\times\frac{1}{3}-\frac{2}{5}\times\frac{1}{6}\)

\(=\frac{2}{5}\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(=\frac{2}{5}\times\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)

\(=\frac{2}{5}\times0\)

\(=0\)

Học tốt

Gọi \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2017}< 1\)

\(=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)

\(=2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2016^2}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)

\(A=1-\frac{1}{2^{2017}}< 1\) (đpcm)

Đặt A=1/2+(1/2)^2+...+(1/2)^2017

=>1/2 A=(1/2)^2+(1/2)^3+...+(1/2)^2017+(1/2)2018      (Nhân cả 2 vế cho 1/2)

=>1/2 A - A=(1/2)^2018-1/2

=>-1/2 A =(1/2)^2018-1/2

=>A=1-(1/2)^2017 <1  (Vì (1/2)^2017>0)

Đug ko biết

\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)

Vậy \(x=-\dfrac{2}{5}\)

x= \(\dfrac{7\pm\sqrt{37}}{3}\) nha

x = 2/5

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