Tính bằng cách thuận tiện
a, \(1\frac{1}{24}\)x\(5\frac{2}{5}\)x\(3\frac{7}{9}\) x\(2\frac{2}{17}\)
b,\(2\frac{3}{13}\) x \(\frac{26}{58}\)x 4 x \(2\frac{15}{24}\)x \(\frac{8}{21}\)
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\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)
= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)
= \(x^8.\frac{1}{10}.1.1.1.1\)
= \(x^8.\frac{1}{10}\)
Mk ko pik co dung ko nua
\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)
Ta xét 2 trường hợp
\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)
tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian
a. \(1\frac{1}{24}\cdot5\frac{2}{5}\cdot3\frac{7}{9}\cdot2\frac{2}{17}\)
\(=\frac{25}{24}\cdot\frac{27}{5}\cdot\frac{34}{9}\cdot\frac{36}{17}\)
\(=\left(\frac{25}{24}\cdot\frac{27}{5}\right)\cdot\left(\frac{34}{9}\cdot\frac{36}{17}\right)=\frac{5\cdot27}{24}\cdot\frac{2\cdot17\cdot9\cdot4}{9\cdot17}=\frac{5\cdot27}{24}\cdot8\)
\(=\frac{5\cdot27\cdot8}{24}=\frac{5\cdot9\cdot3\cdot8}{24}=\frac{45\cdot24}{24}=45\)
b. \(2\frac{3}{13}\cdot\frac{26}{58}\cdot4\cdot2\frac{15}{24}\cdot\frac{8}{21}\)
\(=\frac{29}{13}\cdot\frac{26}{58}\cdot4\cdot\frac{21}{8}\cdot\frac{8}{21}=\frac{29\cdot13\cdot2}{13\cdot29\cdot2}\cdot4\cdot\frac{21\cdot8}{8\cdot21}=1\cdot4\cdot1\)
\(=4\)