Ta có
\(2017-\left(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+\frac{4}{7}+...+\frac{2017}{2020}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{2}{5}+...+\frac{2017}{2020}\right)\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{2}{5}\right)+...+\left(1-\frac{2017}{2020}\right)\)
\(=\frac{3}{4}+\frac{3}{5}+....+\frac{3}{2020}\)
\(=\frac{3.5}{4.5}+\frac{3.5}{5.5}+\frac{3.5}{6.5}+...+\frac{3.5}{2020.5}\)
\(=3.5\left(\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\right)\)
\(=15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)\)
Thế vào ta có
\(\frac{15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)}{\frac{1}{20}+\frac{1}{25}+...+\frac{1}{10100}}=15\)

Được cập nhật 41 giây trước (17:23)

Ta có :
2017−(14 +25 +36 +47 +...+20172020 )
=(1+1+...+1)−(14 +25 +...+20172020 )
=(1−14 )+(1−25 )+...+(1−20172020 )
=34 +35 +....+32020 
=3.54.5 +3.55.5 +3.56.5 +...+3.52020.5 
=3.5(14.5 +15.5 +16.5 +...+12020.5 )
=15.(1

là luc sắp chết mày yêu ai?

 help me

a) \(x^3-\frac{4}{25}x=0\)

\(\Leftrightarrow x\left(x+\frac{2}{5}\right)\left(x-\frac{2}{5}\right)=0\)

<=> x = 0

Xét 2 trường hợp: 

\(\Leftrightarrow x+\frac{2}{5}=0\)

      \(x=0-\frac{2}{5}\)

      \(x=-\frac{2}{5}\)

\(\Leftrightarrow x-\frac{2}{5}=0\)

      \(x=0+\frac{2}{5}\)

      \(x=\frac{2}{5}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{2}{5}\end{cases}}\)

b) \(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{4}{3}\)

\(=\frac{13}{40}:\frac{4}{3}\)

\(=\frac{39}{120}=\frac{13}{40}\)

c) \(4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^2+3\left(\frac{-1}{2}\right)-1\left(\frac{-1}{2}\right)^0\)

\(=4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^3+3\left(\frac{-1}{2}\right)-1.1\)

\(=-\frac{1}{2}-\frac{1}{2}-\frac{3}{2}-1.1\)

\(=-\frac{5}{2}-1\)

\(=-\frac{7}{2}\)

Ta có : 

\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)

\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)

\(=\)\(\frac{2}{7}-\frac{1}{\frac{7}{2}}\)

\(=\)\(\frac{2}{7}-\frac{2}{7}\)

\(=\)\(0\)

Chúc bạn học tốt ~ 

thank nha

k cho mk nhé .OK

 o0oNguyễno0o bn giúp mk nha bài này khó wa

 Harry Potter05 giúp mk đc ko vậy bn!!!

 o0oNguyễno0o giúp mk nhanh ik nha

1/3xD=1/(2x4)+1/(4x6)+...+1/(98x100)
2/3xD=2/(2x4)+2/(4x6)+...+1/(98x100)
2/3xD= 1/2-1/4+1/4-1/6+...+1/98-1/100
2/3xD=1/2-1/100
2/3xD=49/100
D=147/200

 

A=1/2+1/3+1/6+1/12
A=(1/2+1/3+1/6)+1/12
A=1+1/12
A=13/12

B=20/3+61/8+10/3+37/8
B=30/3+98/8
B=10+49/4
B=89/4

\(\frac{2}{7}< \frac{x}{3}< \frac{11}{4};x\inℕ\)

=>\(\frac{12.2}{84}< \frac{28x}{84}< \frac{11.21}{84}\)

=>\(\frac{24}{84}< \frac{28x}{84}< \frac{231}{84}\)

=>24<28x<231

=>28x\(\in\){25;26;27;28;.............................;230}

=>Các số chia hết cho 28 là:28;56;84;112;140;168;196;224

=>x (thỏa mãn)\(\in\){1;2;3;4;5;6;7;8}

Vậy x\(\in\) {1;2;3;4;5;6;7;8}

\(\left(4,5m-\frac{3}{4}.5\frac{1}{3}\right).\frac{1}{12}+\frac{1}{2}x=1\frac{1}{2}\)

\(\left(4,5m-\frac{3}{4}.\frac{16}{3}\right).\frac{1}{2}.\frac{1}{6}+\frac{1}{2}x=\frac{3}{2}\)

\(\left(4,5m-\frac{48}{12}\right).\frac{1}{2}.\left(\frac{1}{6}+x\right)=\frac{3}{2}\)

\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}:\frac{1}{2}\)

\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}.\frac{2}{1}\)

\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{6}{2}\)

\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=3\)

=>3\(⋮\)\(\frac{1}{6}+x\)

=>\(\frac{1}{6}+x\)\(\in\)Ư(3)={\(\pm\)1;\(\pm\)3}

Ta có bảng:

Vậy x\(\in\){\(-1\frac{1}{6}\);\(1\frac{1}{6}\);\(-3\frac{1}{6}\);\(\frac{1}{6}\)}

Chúc bn học tốt

thanks

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}\)

=1

Ta có:

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left[\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right]}=\frac{1}{4}\)

Lại có:

\(\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{3\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}{4\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}=\frac{3}{4}\)

Vậy: 

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{1}{4}+\frac{3}{4}=1\)