A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)

\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)

\(=\frac{3}{5}+\frac{2}{5}=1\)

b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)

 \(=\frac{1}{3.2}-\frac{5.2}{7.3}\)

\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)

\(=\frac{7}{42}-\frac{20}{42}\)

\(=-\frac{13}{42}\)

cs ng làm đung r

đag định lm

#)Giải :

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)

\(=0\)

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)

=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)

\(=0\)

Bài 3:

a, Đặt \(A=\left|2x-\frac{1}{5}\right|+2017\)

Để A đạt GTNN thì \(\left|2x-\frac{1}{5}\right|\)đạt GTNN

Mà \(\left|2x-\frac{1}{5}\right|\ge0\)

Do đó \(\left|2x-\frac{1}{5}\right|=0\)thì A đạt GTNN tức là A = 0 + 2017 = 2017 khi

\(2x-\frac{1}{5}=0=>2x=0+\frac{1}{5}=\frac{1}{5}=>x=\frac{1}{5}.\frac{1}{2}=\frac{1}{10}\)

b, Đặt \(B=\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|\)

Ta thấy \(\frac{1}{2}>\frac{1}{3}>\frac{1}{4}=>x+\frac{1}{2}>x+\frac{1}{3}>x+\frac{1}{4}\)

Do đó để B đạt GTNN thì \(x+\frac{1}{2}\)đạt GTNN

mà \(x+\frac{1}{2}\ge0\)

Từ 2 điều trên => \(x+\frac{1}{2}=0=>x=-\frac{1}{2}\)

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

và \(x+\frac{1}{4}=-\frac{1}{2}+\frac{1}{4}=-\frac{1}{4}\)

Vậy GTNN của \(B=\left|0\right|+\left|-\frac{1}{6}\right|+\left|-\frac{1}{4}\right|=0+\frac{1}{6}+\frac{1}{4}=\frac{10}{24}\)khi x = -1/2

Phần b này thì mình không chắc lắm bạn tự xem lại nhé

Bài 1: 

\(M=\frac{2017}{11-x}\)đạt GTLN <=> 11 - x đạt GTNN và 11 - x > 0 (nếu không thì M đạt giá trị âm (vô lí))

=> 11 - x = 1

=> x = 10

Vậy x = 10 thì M đạt GTLN tức là bằng \(\frac{2017}{1}=2017\)

Bài 2

a, Đặt \(A=-2\left|x-\frac{3}{4}\right|-\left|y+\frac{3}{4}\right|+\frac{5}{6}\)

Để A đạt GTLN <=> \(-2\left|x-\frac{3}{4}\right|\)đạt GTLN và \(\left|y+\frac{3}{4}\right|\)đạt GTNN

mà \(\left|x-\frac{3}{4}\right|\ge0=>-2\left|x-\frac{3}{4}\right|\le0\)

và \(\left|y+\frac{3}{4}\right|\ge0\)

Do đó \(-2\left|x-\frac{3}{4}\right|=0\)và \(\left|y+\frac{3}{4}\right|=0\)

Vậy GTLN của A = 0 - 0 + 5/6 = 5/6 khi

\(\left|x-\frac{3}{4}\right|=0=>x-\frac{3}{4}=0=>x=\frac{3}{4}\)

Và \(\left|y+\frac{3}{4}\right|=0=>y+\frac{3}{4}=0=>y=-\frac{3}{4}\)

b, Đặt \(B=-\left(x+\frac{1}{2}\right)^2+\frac{5}{7}\)

Để B đạt GTLN thì \(-\left(x+\frac{1}{2}\right)^2\)đạt GTLN

Mà \(\left(x+\frac{1}{2}\right)^2\ge0=>-\left(x+\frac{1}{2}\right)^2\le0\)

Do đó để B đạt GTLN thì \(-\left(x+\frac{1}{2}\right)^2=0\)

Khi đó GTLN của B = 0 + 5/7 = 5/7 khi

\(\left(x+\frac{1}{2}\right)^2=0=>x+\frac{1}{2}=0=>x=-\frac{1}{2}\)

\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)

\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)

\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)

2.

\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)

\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)

3.

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

\(=\frac{1}{100}\)

4.

\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)

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

mình chỉ làm được câu 3 thôi

có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)

\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)

\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)

\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)

\(=\frac{-1}{100}\)

Câu 3 như kiên nhé!! nhầm tí !! xl!

C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!

=\(\frac{3}{4}-\frac{5}{2}-\frac{5}{3}+\left(\frac{-1}{8}\right)=\frac{18}{24}-\frac{60}{24}-\frac{40}{24}-\frac{3}{24}=\frac{-85}{24}\)

\(\frac{3}{4}-\left(\frac{5}{2}+\frac{5}{3}\right)+\left(\frac{-1}{2}\right)^3\)

\(=\frac{3}{4}-\frac{25}{6}+\frac{-1}{8}\)

\(=\frac{18-100-3}{24}\)

\(=\frac{-85}{24}\)

Tk mk nha!

Ta có: \(D=2016\left(1-\frac{2}{3}\right)\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{2017}\right)\)

\(=2016.\frac{1}{3}.\frac{3}{5}.\frac{5}{7}...\frac{2015}{2017}\)\(=2016.\left(\frac{1}{3}.\frac{3}{5}.\frac{5}{7}...\frac{2015}{2017}\right)\)

\(=2016\left(\frac{1.3.5.7...2015}{3.5.7....2015.2017}\right)\)\(=2016.\frac{1}{2017}=\frac{2016}{2017}\)

                                                        Vậy \(D=\frac{2016}{2017}\)

sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)

\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)