Bài 1: Thực hiện phép tính( bằng cahcs hợp lí nếu có thể)
\(a)\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-1\frac{4}{13}\)
\(b)12-8.\left(\frac{3}{2}\right)^3\)
\(c)100\times\sqrt{0,04}+\sqrt{\frac{25}{16}}\)
\(d)\frac{1}{4}+\frac{3}{4}.\frac{5}{6}\)
\(e)3,75.7,2+2,8.3,75\)
\(f)\frac{1}{2}\sqrt{4}-\sqrt{25}\)
\(g)(\frac{1}{9})^{2005}.9^{2005}-96^2:24^2\)
\(h)\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)
\(k)\left(-3,5\right).\left(-7,2\right)+\left(-3,15\right).12,4+4,8.\left(-3,15\right)\)
\(l)16\frac{2}{7}:\left(-\frac{2}{5}\right)-28\frac{2}{7}:\left(-\frac{2}{5}\right)\)
\(m)\left(-\frac{1}{2}\right)^3+\frac{1}{2}:5\)
\(n)1\frac{1}{23}+\frac{2}{21}-\frac{1}{23}+\frac{19}{21}+2013^0\)
\(h)\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(i)\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}+1^{2012}}\)
Giải Hộ Mk vs Mk đng cần gấp để nộp
HELP MEEEEEEEEEEEEEEEEEEEE
a) Ta có: \(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+\frac{1}{2}\cdot8-5+64\)
\(=-8+4-5+64=55\)
b) Ta có: \(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{3}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}\right)\cdot\frac{3}{2}+\left(\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=0\cdot\frac{3}{2}=0\)
c) Ta có: \(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\frac{2^{10}\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}=\frac{2\left(2^9\cdot9^4-6^9\right)}{6^8\left(2^2+20\right)}=\frac{-1}{3}\)
a) ( -2 )3 + \(\frac{1}{2}:\frac{1}{8}\) - √25 + \(|-64|\)
= \(\frac{-8}{1}\) + \(\frac{1}{2}.\frac{8}{1}\) - \(\frac{5}{1}\) + \(\frac{64}{1}\)
= \(\frac{-16}{2}+\frac{1}{2}.\frac{8}{1}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16}{2}+\frac{8}{2}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16+8-10+128}{2}\) = \(\frac{110}{2}\) = 55