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Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
a) \(x^3-\frac{4}{25}x=0\)
\(\Leftrightarrow x\left(x+\frac{2}{5}\right)\left(x-\frac{2}{5}\right)=0\)
<=> x = 0
Xét 2 trường hợp:
\(\Leftrightarrow x+\frac{2}{5}=0\)
\(x=0-\frac{2}{5}\)
\(x=-\frac{2}{5}\)
\(\Leftrightarrow x-\frac{2}{5}=0\)
\(x=0+\frac{2}{5}\)
\(x=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{2}{5}\end{cases}}\)
b) \(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{4}{3}\)
\(=\frac{13}{40}:\frac{4}{3}\)
\(=\frac{39}{120}=\frac{13}{40}\)
c) \(4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^2+3\left(\frac{-1}{2}\right)-1\left(\frac{-1}{2}\right)^0\)
\(=4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^3+3\left(\frac{-1}{2}\right)-1.1\)
\(=-\frac{1}{2}-\frac{1}{2}-\frac{3}{2}-1.1\)
\(=-\frac{5}{2}-1\)
\(=-\frac{7}{2}\)