\(B=\frac{2}{15}+\left(\frac{5}{9}+\frac{-6}{9}\right)\)

\(\Leftrightarrow B=\frac{2}{15}+\frac{-1}{9}\)

\(\Leftrightarrow B=\frac{12}{90}+\frac{-10}{90}\)

\(\Leftrightarrow B=\frac{2}{90}\)

\(\Leftrightarrow B=\frac{1}{45}\)

\(B=\frac{2}{15}=\left(\frac{5}{9}+\frac{-6}{9}\right)\)

\(B=\frac{2}{15}+\frac{-1}{9}\)

\(B=\frac{18}{135}+\frac{-15}{135}\)

\(B=\frac{3}{135}\)

\(B=\frac{2}{15}+\left(\frac{5}{9}+-\frac{6}{9}\right)\)

\(B=\frac{2}{15}+-\frac{1}{9}\)

\(B=\frac{1}{45}\)

\(Linh\)\(Yukiko\)

=\(\frac{36-4+3}{6}-\frac{30+10-9}{6}-\frac{18-14+15}{6}\)

=\(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{35-31-19}{6}-\frac{15}{6}=-\frac{5}{2}\)

bài này thì dễ:

\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

Cách 1:

\(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{35-31-19}{6}=-\frac{15}{6}=-\frac{5}{2}\)

Cách 2: 

\(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}=-2\frac{1}{2}=-\frac{5}{2}\)

Cách 1 :

\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

\(=\left(\frac{36-4+3}{6}\right)-\left(\frac{30+10-9}{6}\right)-\left(\frac{18-14+15}{8}\right)\) 

\(=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=-\frac{15}{6}=-\frac{5}{2}\)

Cách 2 : \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

\(=\left(6-5-3\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)\)

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

kết bạn nhé

bn gửi nhé

Tổng quát: a^4+1/4=(a^2+1/2)^2-a^2=(a^2+1/2-a)(a^2+1/2+a)=[(a-1/2)^2+1/4][(a^1/2)^2+1/4]=[(a-0,5)^2+0,25][(a+0,5)^2+0,25]

Tử số của M=[(2-0,5)^2+0,25][(2+0,5)^2+0,25][(4-0,5)^2+0,25][(4+0,5)^2+0,25][(6-0,5)^2+0,25][(6+0,5)^2+0,25]....[(2014-0,5)^2+0,25][(2014+0,5)^2+0,25]

                  =(1,5^2+0,25)(2,5^2+0,25)(3,5^2+0,25)(4,5^2+0,25)(5,5^2+0,25)(6,5^2+0,25)....(2013,5^2+0,25)(2014,5^2+0,25)

Mẫu số của M=[(1-0,5)^2+0,25][(1+0,5)^2+0,25][(3-0,5)^2+0,25][(3+0,5)^2+0,25][(5-0,5)^2+0,25][(5+0,5)^2+0,25]....[(2013-0,5)^2+0,25][(2013+0,5)^2+0,25]

                    =(0,5^2+0,25)(1,5^2+0,25)(2,5^2+0,25)(3,5^2+0,25)(4,5^2+0,25)(5,5^2+0,25)....(2012,5^2+0,25)(2013,5^2+0,25)

Vậy M=(2014,5^2+0,25)/(0,5^2+0,25)

Còn bao nhiêu bạn tính tiếp nhá

\(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\)

Nhận thấy:  \(\left|2x+1\right|\ge0\);     \(\left|x+y-\frac{1}{2}\right|\ge0\)

=>   \(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\ge0\)

Dấu "=" xảy ra  <=>  \(\hept{\begin{cases}2x+1=0\\x+y-\frac{1}{2}=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=-\frac{1}{2}\\y=1\end{cases}}\)

đến đây bạn thay x,y tìm đc vào A để tính nhé

Xét số hạng tổng quát: 

\(k^4+\frac{1}{4}=\left(k^4+2\cdot\frac{1}{2}\cdot k^2+\frac{1}{4}\right)-k^2\)

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

Thay k từ 1 đến 2014 , ta được

M=

\(\frac{\left(2+\frac{1}{2}\right)\left(6+\frac{1.}{2}\right)...\left(4054182+\frac{1}{2}\right)\left(4058210+\frac{1}{2}\right)}{\frac{1}{2}\cdot\left(2+\frac{1}{2}\right)...\left(4050156+\frac{1}{2}\right)\left(4054182+\frac{1}{2}\right)}\)=\(\frac{4058210+\frac{1}{2}}{\frac{1}{2}}=8116421\)

\(M=\frac{5}{7}\left(\frac{8}{5}+\frac{2}{5}\right)+\frac{5}{7}\left(\frac{6}{5}+\frac{9}{5}\right)\)

Cách 1 : \(M=\frac{5}{7}\left(\frac{8}{5}+\frac{2}{5}\right)+\frac{5}{7}\left(\frac{6}{5}+\frac{9}{5}\right)\)

\(M=\frac{5}{7}\cdot\left(\frac{8}{5}+\frac{2}{5}+\frac{6}{5}+\frac{9}{5}\right)=\frac{5}{7}.\frac{25}{5}=\frac{25}{7}\)

Cách 2:

\(M=\frac{5}{7}\left(\frac{8}{5}+\frac{2}{5}\right)+\frac{5}{7}\left(\frac{6}{5}+\frac{9}{5}\right)\)

\(M=\frac{10}{5}.\frac{5}{7}+\frac{15}{5}.\frac{5}{7}=\frac{10}{7}+\frac{15}{7}=\frac{25}{7}\)

Cách 3 :

\(M=\frac{5}{7}\left(\frac{8}{5}+\frac{2}{5}\right)+\frac{5}{7}\left(\frac{6}{5}+\frac{9}{5}\right)\)

\(M=\frac{8}{5}.\frac{5}{7}+\frac{2}{5}.\frac{5}{7}+\frac{6}{5}.\frac{5}{7}+\frac{9}{5}.\frac{5}{7}=\frac{8}{7}+\frac{2}{7}+\frac{6}{7}+\frac{9}{7}=\frac{25}{7}\)

Cách 1 :

\(M=\left(\frac{8}{5}+\frac{2}{5}\right).\frac{5}{7}+\left(\frac{6}{5}+\frac{9}{5}\right).\frac{5}{7}\)

\(M=2.\frac{5}{7}+3.\frac{5}{7}\)

\(M=\frac{10}{7}+\frac{15}{7}\)

\(M=\frac{25}{7}\)

Cách 2

\(M=\left(\frac{8}{5}+\frac{2}{5}\right).\frac{5}{7}+\left(\frac{6}{5}+\frac{9}{5}\right).\frac{5}{7}\)

\(M=2.\frac{5}{7}+3.\frac{5}{7}\)

\(M=\left(2+3\right).\frac{5}{7}\)

\(M=5.\frac{5}{7}\)

\(M=\frac{25}{7}\)

Cách 3 :

\(M=\left(\frac{8}{5}+\frac{2}{5}\right).\frac{5}{7}+\left(\frac{6}{5}+\frac{9}{5}\right).\frac{5}{7}\)

\(M=\left[\left(\frac{8}{5}+\frac{2}{5}\right)+\left(\frac{6}{5}+\frac{9}{5}\right)\right].\frac{5}{7}\)

\(M=\left[2+3\right].\frac{5}{7}\)

\(M=5.\frac{5}{7}\)

\(M=\frac{25}{7}\)

Biết có vậy thôi

có thể có 4 cách đấy