Tính giâ trị của biểu thức sau theo công thức sau theo nhiều cách khác nhau :
\(M=\left(\frac{8}{5}+\frac{2}{5}\right).\frac{5}{7}+\left(\frac{6}{5}+\frac{9}{5}\right).\frac{5}{7}\)
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\(B=\frac{2}{15}+\left(\frac{5}{9}+\frac{-6}{9}\right)\)
\(\Leftrightarrow B=\frac{2}{15}+\frac{-1}{9}\)
\(\Leftrightarrow B=\frac{12}{90}+\frac{-10}{90}\)
\(\Leftrightarrow B=\frac{2}{90}\)
\(\Leftrightarrow B=\frac{1}{45}\)
=\(\frac{36-4+3}{6}-\frac{30+10-9}{6}-\frac{18-14+15}{6}\)
=\(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{35-31-19}{6}-\frac{15}{6}=-\frac{5}{2}\)
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:
\(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}=-2\frac{1}{2}=-\frac{5}{2}\)
\(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\)
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
Cách 1 : \(M=\frac{10}{5}.\frac{5}{7}+\frac{15}{5}.\frac{5}{7}=\frac{10}{7}+\frac{15}{7}=\frac{25}{7}\)
Cách 2 : \(M=\frac{8}{5}.\frac{5}{7}+\frac{2}{5}.\frac{5}{7}+\frac{6}{5}.\frac{5}{7}+\frac{9}{5}.\frac{5}{7}\)
\(=\frac{8}{7}+\frac{2}{7}+\frac{6}{7}+\frac{9}{7}=\frac{25}{7}\)
Cách 3 : \(M=\left(\frac{8}{5}+\frac{2}{5}+\frac{6}{5}+\frac{9}{5}\right).\frac{5}{7}=\frac{25}{5}.\frac{5}{7}=\frac{25}{7}\)