27 Tính
A= \(\dfrac{10\dfrac{1}{3}.(26\dfrac{1}{3}-\dfrac{176}{7})-\dfrac{12}{11}.(\dfrac{10}{3}-1,75}{(\dfrac{5}{91-0,25)}.\dfrac{60}{11}-1}\)
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\(=\dfrac{\dfrac{31}{3}\cdot\left(26+\dfrac{1}{3}-\dfrac{176}{7}\right)-\dfrac{12}{11}\cdot\left(\dfrac{10}{3}-\dfrac{7}{4}\right)}{\left(\dfrac{5}{90.75}\right)\cdot\dfrac{60}{11}-1}\)
\(=\dfrac{\dfrac{31}{3}\cdot\dfrac{25}{21}-\dfrac{12}{11}\cdot\dfrac{19}{12}}{-\dfrac{931}{1331}}\)
\(=\dfrac{7328}{693}:\dfrac{-931}{1331}\simeq-15.118\)
1: \(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
=5/13+3
=5/13+39/13
=44/13
2: \(B=\dfrac{2^{15}\cdot3^{30}\cdot5-5\cdot2^5\cdot7^5\cdot2^{12}}{3^3\cdot2\cdot2^{14}\cdot3^{14}\cdot3^{14}-2^2\cdot3\cdot2^{15}\cdot7^5}\)
\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3^{31}\cdot2^{15}-2^{17}\cdot3\cdot7^5}\)
\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3\cdot2^{15}\cdot\left(3^{30}-2^2\cdot7^5\right)}=\dfrac{5}{3}\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
\(A=\dfrac{115-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{30}}\)
\(=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
\(=\dfrac{5}{13}+3\)
\(=\dfrac{5}{13}+\dfrac{39}{13}\)
\(=\dfrac{44}{13}\)
\(=3\dfrac{5}{13}\)
CHÚC BN HC TỐT
\(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
=5/13+3
=39/13+5/13=44/13
3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)
c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)
d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)
a: \(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\left(\dfrac{1}{5}-\dfrac{3}{10}+\dfrac{1}{13}\right)}\)
=5/13+3
=5/13+39/13
=44/13
b: \(B=\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^5\cdot7^5\cdot2^{12}}{2\cdot3^3\cdot2^{14}\cdot3^{14}\cdot3^{14}-3\cdot2^2\cdot2^{15}\cdot7^5}\)
\(=\dfrac{2^{15}\cdot5\left(3^{30}-7^5\cdot2^2\right)}{2^{15}\cdot3^{31}-3\cdot2^{17}\cdot7^5}\)
\(=\dfrac{2^{15}\cdot5\left(3^{30}-7^5\cdot2^2\right)}{2^{15}\cdot3\left(3^{30}-7^5\cdot2^2\right)}=\dfrac{5}{3}\)