(-2/7)^2-(2/7)^3 NÊU CÁCH LÀM
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\(\frac{x}{5}=\frac{4,2}{8,4}\)
\(\frac{x}{5}=\frac{1}{2}\)
\(\Leftrightarrow\)\(2x=5\)
\(\Leftrightarrow\)\(x=\frac{5}{2}\)
Từ \(2x=3y\)\(\Rightarrow\)\(\frac{x}{3}=\frac{y}{2}=\frac{x}{3}.\frac{1}{7}=\frac{y}{2}.\frac{1}{7}\)\(\Rightarrow\)\(\frac{x}{21}=\frac{y}{14}\)( 1 )
Từ \(5y=7z\)\(\Rightarrow\)\(\frac{y}{7}=\frac{z}{5}=\frac{y}{7}.\frac{1}{2}=\frac{z}{5}.\frac{1}{2}\)\(\Rightarrow\)\(\frac{y}{12}=\frac{z}{10}\)( 2 )
Từ ( 1 ) và ( 2 ) suy ra : \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)
Đặt \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=k\)\(\Rightarrow\hept{\begin{cases}x=21k\\y=14k\\z=10k\end{cases}}\)
Thay vào ta được :
\(2.21k-3.14k+4.10k=350\)
\(42k-42k+40k=350\)
\(40k=350\)
\(k=\frac{35}{4}\)
\(\Rightarrow\hept{\begin{cases}x=21.\frac{35}{4}\\y=14.\frac{35}{4}\\z=10.\frac{35}{4}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{735}{4}\\y=\frac{245}{2}\\z=\frac{175}{2}\end{cases}}\)
Ta có :
( x + 0,8 )2 = 0,25
\(\Rightarrow\left(x+\frac{4}{5}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}\left(x+\frac{4}{5}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x+\frac{4}{5}\right)^2=\left(\frac{-1}{2}\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{4}{5}=\frac{1}{2}\\x+\frac{4}{5}=\frac{-1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{10}\\x=\frac{-13}{10}\end{cases}}\)
\(\left(x+0,8\right)^2=0,25\)
\(\left(x+0,8\right)^2=\frac{1}{4}\)
\(\left(x+0,8\right)^2=\left(\frac{1}{2}\right)^2\)hoặc \(\left(-\frac{1}{2}\right)^2\)
\(x+0,8=\pm\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x+0,8=\frac{1}{2}\\x+0,8=-\frac{1}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}-0,8\\x=-\frac{1}{2}-0,8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{10}\\x=-\frac{13}{10}\end{cases}}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Lại có \(\frac{30}{26}+\frac{31}{27}+...+\frac{54}{50}-25\)
\(=\left(\frac{30}{26}-1\right)+\left(\frac{31}{27}-1\right)+...+\left(\frac{54}{50}-1\right)\)
\(=\frac{4}{26}+\frac{4}{27}+...+\frac{4}{50}=4\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)\)
Khi đó M = \(\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\right):\left[4\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)\right]=\frac{1}{4}\)