\(\frac{1}{2}-\left(\frac{2}{3}x-\frac{1}{3}\right)=\frac{2}{3}\)

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

\(\frac{2}{3}x-\frac{1}{3}=\frac{-1}{6}\)

\(\frac{2}{3}x=\frac{-1}{6}+\frac{1}{3}\)

\(\frac{2}{3}x=\frac{1}{6}\)

\(x=\frac{1}{6}:\frac{2}{3}\)

\(x=\frac{1}{4}\)

~ Hok tốt ~

\(\frac{3}{x+5}=15\%\)

\(\Leftrightarrow\frac{3}{x+5}=\frac{15}{100}\)

\(\Leftrightarrow\frac{3}{x+5}=\frac{3}{20}\)

\(\Leftrightarrow x+5=20\)

\(\Leftrightarrow x=20-5\)

\(\Leftrightarrow x=15\)

Giúp mình nha

x + 25 = 64

x         = 64 - 25

x         = 39

Vậy x = 39

A =  1 + 2 + 3 + ... + 2018 (có 2018 số )

   = (2018 + 1) . 2018 : 2 = 2037171

B = 1 + 3 + 5 + ... + 2017(có  1009 số )

   = (2017 + 1) . 1009 : 2 = 1018081

C = 2 + 4 + 6 + ... + 2018 (Có 1009 số )

   = (2018 + 2) x 1009 : 2 = 1019090

D = 72 . 153 + 27.153 + 153

    = (72 + 27 + 1) . 153

    = 100 . 153 = 15300

nhân 3 vào mỗi hạng tử ta được:

3*(1.2+2.3+3.4+...+99.100)

= 1.2.(3-0)+ 2.3.(4-1)+ 3.4.(5-2)+... + 99.100.(101-98)

=1.2.3 + 2.3.4 -1.2.3 + 3.4.5 -2.3.4 +... + 99.100.101 - 98.99.100

= 99.100.101

Vậy tổng ban đầu 99.100.101/3= 33.100.101

Vậy tổng trên chia hết cho 2;3;4;5;10

à có ai chơi ngọc rồng không cho mk 1 nick có ddeeej là được

\(A=\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+...+\frac{3}{49\cdot50}\)

\(A=3\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\right)\)

\(A=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A=3\left(1-\frac{1}{50}\right)\)

\(A=3\cdot\frac{49}{50}=\frac{147}{50}\)

A = 3/1.2 + 3/2.3 + 3/3.4 + ... + 3/49.50     

A=3/1-3/2+3/2-3/3+3/3-3/4+...+3/49-3/50

A=3/1-3/50

A= 147/50

a ) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

Vi \(1-\frac{1}{50}< 1\)

=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}< 1\)

b ) Dat B = \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2013^2}\)

Ta co :

  \(\frac{1}{5^2}< \frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)

\(\frac{1}{6^2}< \frac{1}{5.6}=\frac{1}{5}-\frac{1}{6}\)

\(\frac{1}{7^2}< \frac{1}{6.7}=\frac{1}{6}-\frac{1}{7}\)

...

\(\frac{1}{2013^2}< \frac{1}{2012.2013}=\frac{1}{2012}-\frac{1}{2013}\)

Vay \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2013^2}< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2012}-\frac{1}{2013}\)

=> B < \(\frac{1}{4}-\frac{1}{2013}\)

Ma \(\frac{1}{4}-\frac{1}{2013}< \frac{1}{4}\)

=> B < \(\frac{1}{4}\)

KL : \(Vay\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2013^2}< \frac{1}{4}\)

Cho Câu hỏi

Cho đáp án

Chứ bây giờ ở đây mk ko có đề bạn nhé

Chắc muốn nhận vé bá cáo quá haa :))))))))))

\(5^{21}\over4\) (ko biết có đúng ko)

\(S=1+5+5^2+5^3+...+5^{20}\)

\(\Rightarrow5S=5+5^2+5^3+5^4+...+5^{21}\)

\(\Rightarrow5S-S=\left(5+5^2+5^3+...+5^{21}\right)-\left(1+5+5^2+...+5^{20}\right)\)

\(\Rightarrow4S=5^{21}-1\)

\(\Rightarrow S=\frac{5^{21}-1}{4}\)

vãi hỏi đáp án mà cách làm bt lâu rồi