Vũ Minh Tuấn giúp mình vs

Lời giải:

$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+......+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$

$4D=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$

Trừ theo vế:

\(3D=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)

\(\Rightarrow 12D=4+1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2017}}-\frac{2019}{4^{2018}}\)

Trừ theo vế:
$9D=4-\frac{2019}{4^{2018}}+\frac{2019}{4^{2019}}-\frac{1}{4^{2018}}$

$=4-\frac{6061}{4^{2019}}< 4$

$\Rightarrow D< \frac{4}{9}<\frac{4}{8}$ hay $D< \frac{1}{2}$ (đpcm)

\(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}\)

\(\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)

\(\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)

\(-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}\)

\(\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}\)

Đặt \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)

\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)

\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)

\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)

\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)

\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)

\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)

\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)

\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)

B= \(\frac{2\cdot2018}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}}\)

Ta có: 

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}\)

\(=1+\frac{1}{\frac{3.2}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{2018.2019}{2}}\)

\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2018.2019}\)

\(=\frac{2}{2}+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)

\(=2\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=2\left(1-\frac{1}{2019}\right)=\frac{2.2018}{2019}\)

=> B= \(\frac{2\cdot2018}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}}=\frac{2.2018}{\frac{2.2018}{2019}}=2019\)

\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

help me

\(a)\) Ta có : 

\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)

\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)

\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)

\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

Lại có : 

\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)

\(\Rightarrow\)\(x=2019\)

Vậy \(x=2019\)

Chúc bạn học tốt ~ 

\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)

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

\(\Leftrightarrow\)\(x+1=2019\)

\(\Leftrightarrow\)\(x=2019-1\)

\(\Leftrightarrow\)\(x=2018\)

Vậy \(x=2018\)

Chúc bạn học tốt ~ 

Đề thi đó

\(\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)\)

Với n thuộc N*, ta có:

\(\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n^2}+\frac{2\left(n+1-n-1\right)}{n\left(n+1\right)}}\)\)

\(\(=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+2.1.\frac{1}{n}-2.1.\frac{1}{n+1}-2.\frac{1}{n}.\frac{1}{\left(n+1\right)}}\)\)

\(\(=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n-1}\right)^2}=1+\frac{1}{n}-\frac{1}{n-1}\)\). Áp dụng vô bài, ta có:

\(\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)\)

\(\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2017}-\frac{1}{2018}\)\)

\(\(=2016+\frac{1}{2}-\frac{1}{2018}=2016\frac{504}{1009}\)\)

P/s: Lại là thằng quỷ Thắng

Xét số hạng tổng quát

 \(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=1^2+\left(\frac{1}{k}\right)^2+\left(\frac{1}{k+1}\right)^2+2.1.\frac{1}{k}-2.\left(\frac{1}{k}.\frac{1}{k+1}\right)-2.1.\frac{1}{k+1}\)

\(=\left(1+\frac{1}{k}-\frac{1}{k+1}\right)^2\)

( Vì \(\frac{1}{k}-\frac{1}{k\left(k+1\right)}-\frac{1}{k+1}=\frac{k+1-1-k}{k\left(k+1\right)}=0\) )

Vậy thì \(\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=1+\frac{1}{k}-\frac{1}{k+1}\)

Vậy \(A=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)

\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2017}-\frac{1}{2018}\)

\(=2016+\frac{1}{2}-\frac{1}{2018}=2016\frac{504}{1009}\)

tìm mãi mới ra cái link == Câu hỏi của Tô Hoài Dung - Toán lớp 9 - Học toán với OnlineMath