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\(=\left(\dfrac{1}{2}-\dfrac{6}{5}\right):\dfrac{21}{20}-\dfrac{25}{4}+\dfrac{1}{2}\)
\(=\dfrac{-7}{10}\cdot\dfrac{20}{21}-\dfrac{25}{4}+\dfrac{2}{4}\)
\(=\dfrac{-2}{3}-\dfrac{23}{4}=\dfrac{-8-69}{12}=\dfrac{-77}{12}\)
\(\left(\sqrt{\dfrac{1}{4}}-1,2\right):1\dfrac{1}{20}-\left(-\dfrac{5}{2}\right)^2+\left|1,25-\dfrac{3}{4}\right|\)
\(=-\dfrac{7}{10}:\dfrac{21}{20}-\dfrac{25}{4}+\left|\dfrac{1}{2}\right|\)
\(=-\dfrac{7}{10}.\dfrac{20}{21}-\dfrac{25}{4}+\dfrac{1}{2}\)
\(=-\dfrac{2}{3}-\dfrac{25}{4}+\dfrac{1}{2}\)
\(=-\dfrac{77}{12}\)
Tính hợp lí nếu có thể:
a,\(\left(1,2-\sqrt{\dfrac{1}{4}}\right):1\dfrac{1}{20}+\left|\dfrac{3}{4}-1,25\right|-\left(-\dfrac{3}{2}\right)^2\)
\(=\left(\dfrac{12}{10}-\dfrac{1}{2}\right):\dfrac{21}{20}+\left|\dfrac{3}{4}-\dfrac{5}{4}\right|-\dfrac{9}{4}\)
\(=\dfrac{7}{10}\cdot\dfrac{20}{21}+\left|-\dfrac{1}{2}\right|-\dfrac{9}{4}\)
\(=\dfrac{2}{3}+\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{8+6-27}{12}=-\dfrac{13}{12}\)
\(\dfrac{\dfrac{4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}+\left(\dfrac{6}{5}\cdot\dfrac{1}{2}\right):\dfrac{4}{5}\)
\(=\dfrac{4}{5}:\dfrac{3}{5}+\dfrac{7}{4}:7+\dfrac{3}{5}:\dfrac{4}{5}\)
\(=\dfrac{4}{3}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\dfrac{7}{3}\)
e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
\(=\dfrac{0,8:1}{0,6}+\dfrac{1:\dfrac{4}{7}}{3\dfrac{11}{36}.2\dfrac{2}{17}}+0,6:\dfrac{4}{5}\)
\(=\dfrac{0,4}{0,3}+\dfrac{\dfrac{7}{4}}{7}+0,75\)
\(=1\dfrac{0,4}{0,3}\)
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
\(=\left(\dfrac{1}{2}-\dfrac{6}{5}\right):\dfrac{21}{20}-\dfrac{25}{4}+\dfrac{1}{2}=\dfrac{-7}{10}\cdot\dfrac{20}{21}-\dfrac{23}{4}\)
\(=\dfrac{-2}{3}-\dfrac{23}{4}=\dfrac{-8-69}{12}=-\dfrac{77}{12}\)