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a: \(=\dfrac{-3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)=\dfrac{-3}{4}\cdot40=-30\)
b: \(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)
\(=\dfrac{35}{6}:\dfrac{-175+132}{42}+\dfrac{15}{2}\)
\(=\dfrac{35}{6}\cdot\dfrac{42}{-43}+\dfrac{15}{2}\)
\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}\)
\(=\dfrac{-70\cdot7+15\cdot43}{86}=\dfrac{155}{86}\)
c: \(=\dfrac{-7}{5}\left(4+\dfrac{5}{9}+5+\dfrac{4}{9}\right)=\dfrac{-7}{5}\cdot10=-14\)
d: \(=4+\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}\cdot\dfrac{64}{125}\cdot\dfrac{-8}{27}\right)\)
\(=\dfrac{89}{16}+25\cdot\dfrac{-32}{375}\)
\(=\dfrac{89}{16}-\dfrac{32}{15}=\dfrac{823}{240}\)
e: \(=\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
1: \(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{3}-\dfrac{1}{4}\cdot\dfrac{5}{6}=\dfrac{3}{4}+\dfrac{10}{3}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{80}{24}-\dfrac{5}{24}=\dfrac{93}{24}=\dfrac{31}{8}\)
2: \(=\left(7+\dfrac{23}{27}-\dfrac{23}{27}\right)+\left(\dfrac{11}{25}+\dfrac{14}{25}\right)+3.25\)
\(=7+1+3.25=8+3.25=11.25\)
3: \(=\left(\dfrac{1}{9}\cdot9\right)^{2005}-4^2=1-16=-15\)
4: \(=2\cdot\dfrac{9}{4}-\dfrac{7}{2}=\dfrac{9}{2}-\dfrac{7}{2}=1\)
5: \(=\dfrac{15}{2}\cdot\dfrac{-3}{5}+\dfrac{5}{2}\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\cdot\left(\dfrac{15}{2}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot10=-6\)
6: \(=\left(\dfrac{6}{10}+\dfrac{5}{10}\right)^2=\left(\dfrac{11}{10}\right)^2=\dfrac{121}{100}\)
7: \(=\dfrac{1}{2}\cdot\dfrac{-7}{2}=\dfrac{-7}{4}\)
a, \(-\frac{187}{70}\)
b,\(\frac{27}{70}\)
c,\(\frac{53}{14}\)
d,\(\frac{27}{4}\)
e,1
f,\(\frac{23}{4}\)
g,-1
i,6
k,315
l,\(\frac{9}{2}\)
\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)
\(\Rightarrow\dfrac{x-1}{50}-1+\dfrac{x-2}{49}-1=\dfrac{x-3}{48}-1+\dfrac{x-4}{47}-1\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}=\dfrac{x-51}{48}+\dfrac{x-51}{47}\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}-\dfrac{x-51}{48}-\dfrac{x-51}{47}=0\)
\(\Rightarrow\left(x-51\right)\left(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\right)=0\)
Vì \(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\ne0\) nên \(x-51=0\Rightarrow x=51\)
\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)
\(\Rightarrow\dfrac{x+25}{6}+1+\dfrac{x+20}{11}+1+\dfrac{x+16}{15}+1=0\)
\(\Rightarrow\dfrac{x+31}{6}+\dfrac{x+31}{11}+\dfrac{x+31}{15}=0\)
\(\Rightarrow\left(x+31\right)\left(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\ne0\) nên \(x+31=0\Rightarrow x=-31\)
\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)
\(\Rightarrow\dfrac{x-15}{6}-1+\dfrac{x-10}{11}-1=\dfrac{x-3}{18}-1+\dfrac{x-7}{14}-1\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}=\dfrac{x-21}{18}+\dfrac{x-21}{14}\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}-\dfrac{x-21}{18}-\dfrac{x-21}{14}=0\)
\(\Rightarrow\left(x-21\right)\left(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\ne0\) nên \(x-21=0\Rightarrow x=21\)
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
\(\sqrt{\frac{16}{25}}\) + (-\(\frac12\))\(^3\) - |- \(\frac{7}{11}\)| x 2\(\frac34\)
= \(\frac45\) - \(\frac18\) - \(\frac{7}{11}\) x \(\frac{11}{4}\)
= \(\frac45-\frac18\) - \(\frac74\)
= \(\frac{32}{40}\) - \(\frac{5}{40}\) - \(\frac{70}{40}\)
= \(\frac{27}{40}\) - \(\frac{70}{40}\)
= - \(\frac{43}{40}\)
what
\(\sqrt{\frac{16}{25}}\) +\(\left(-\frac12\right)^3\) - \(\left|-\frac{7}{11}\right|\) . 2\(\frac34\)
=\(\frac45\) + \(\left(-\frac18\right)\) - \(\frac{7}{11}\) . 2\(\frac34\)
= \(\frac45\) + \(\left(-\frac18\right)\) - \(\frac74\)
= \(-\frac{13}{40}\)
Ta có: \(\sqrt{\frac{16}{25}}+\left(-\frac12\right)^3-\left|-\frac{7}{11}\right|\cdot2\frac34\)
\(=\frac45-\frac18-\frac{7}{11}\cdot\frac{11}{4}\)
\(=\frac45-\frac18-\frac74=\frac45-\frac18-\frac{14}{8}=\frac45-\frac{15}{8}=\frac{32-75}{40}=\frac{-43}{40}\)