a)0,5-|x-3,5|

         Vì |x-3,5|\(\ge0\)

                   Do đó 0,5-|x-3,5|\(\ge0,5\)

 Dấu = xảy ra khi x-3,5=0

                            x=3,5

Vậy Max A=0,5 khi x=3,5

Mỏi cổ quá khi đọc đề bài của bn nên mk làm câu a thôi

     Vậy 
 

c) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}\)

\(=\frac{1.2.3.4...2014}{2.3.4.5...2015}=\frac{\left(1.2.3.4...2014\right)}{\left(2.3.4.5...2014\right).2015}=\frac{1}{2015}\)

lên olm.vn mà giải

Bài 1:

c/ \(\left(2017-\dfrac{5}{181}+\dfrac{1}{50}\right)-\left(4+\dfrac{3}{181}-\dfrac{3}{50}\right)-\left(1-\dfrac{8}{181}+\dfrac{3}{50}\right)\)

\(=2017-\dfrac{5}{181}+\dfrac{1}{50}-4-\dfrac{3}{181}+\dfrac{3}{50}-1+\dfrac{8}{181}-\dfrac{3}{50}\)

\(=2012+\dfrac{1}{50}=2012,02\)

d/ \(1-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}-...-\dfrac{1}{99\cdot100}\)

\(=1-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)

\(=1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=1-\left(1-\dfrac{1}{100}\right)=1-1+\dfrac{1}{100}=\dfrac{1}{100}\)

a) 4126

b) 615

c) 927

b) 615

c) 927

a, \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{27.30}.\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{27}-\dfrac{1}{30}.\)

\(=1+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(\dfrac{1}{7}+\dfrac{1}{7}\right)+...+\left(\dfrac{1}{27}-\dfrac{1}{27}\right)-\dfrac{1}{30}.\)

\(=1+0+0+...+0-\dfrac{1}{30}.\)

\(=1-\dfrac{1}{30}=\dfrac{29}{30}.\)

Vậy \(B=\dfrac{29}{30}.\)

b, \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}.\)

\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+\dfrac{1}{8.8}.\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}.\)

\(< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}.\)

\(< 1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{6}-\dfrac{1}{6}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\dfrac{1}{8}.\)

\(< 1+0+0+0+0+0+0-\dfrac{1}{8}.\)

\(< 1-\dfrac{1}{8}=\dfrac{7}{8}< 1.\)

\(\Rightarrow B< \dfrac{7}{8}< 1.\)

\(\Rightarrow B< 1\left(đpcm\right).\)

~ Chúc bn học tốt!!! ~

Bài mik đúng thì nhớ tick mik nha!!! ^ - ^

bài 5 cơ mak

đù má đẹp ***** cho 1000 LIKE

phải nói là quá đẹp cho mình like nha

A\(=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

A=\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)

A=\(\dfrac{1}{5}-\dfrac{1}{12}\)

A=\(\dfrac{12}{60}-\dfrac{5}{60}\)

A=\(\dfrac{7}{60}\)

\(A=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

\(A=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{11.12}\)

\(A=\dfrac{1}{5}-\dfrac{1}{12}\)

\(A=\dfrac{7}{60}\)

Số cam trong vườn là :

450 . \(\frac{2}{5}\) = 180 ( cây )

Số hồng xiêm trong vườn là :

450 . 50% = 225 ( cây )

Số bưởi trong vườn là :

450 - 180 - 225 = 45 ( cây )

Đ/s : 45 cây bưởi

45 cây bưởi nha bạn!! Mình làm câu này zồi

\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(C=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\left(\frac{1}{2013}-\frac{1}{2015}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)

\(C=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)

mấy bài còn lại dễ nên bạn tự làm nha

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