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\(a,A=\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left(3+\dfrac{5}{6}-1+\dfrac{1}{3}\right)\left(3+\dfrac{4}{15}-2+\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left[\left(3-1\right)+\left(\dfrac{5}{6}+\dfrac{1}{3}\right)\right]+\left[\left(3-2\right)+\left(\dfrac{4}{15}+\dfrac{3}{5}\right)\right]\)
\(\Leftrightarrow A=\left[2+\left(\dfrac{5}{6}+\dfrac{2}{6}\right)\right]+\left[1+\left(\dfrac{4}{15}+\dfrac{9}{15}\right)\right]\)
\(\Leftrightarrow A=\left(2+\dfrac{7}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=\left(2+1+\dfrac{1}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=3\dfrac{1}{6}+1\dfrac{13}{15}\)
Vậy...
\(b,B=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^{10}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(\Leftrightarrow B=\dfrac{6}{\left(2.3\right).5}\)
\(\Leftrightarrow B=\dfrac{6}{6.5}\)
\(\Leftrightarrow B=\dfrac{1}{5}\)
Vậy....
a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)
\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)
\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)
\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)=\dfrac{7}{8}.\left(\dfrac{10}{60}+\dfrac{24}{60}\right)=\dfrac{7}{8}.\dfrac{17}{30}=\dfrac{114}{240}\)
b)\(\dfrac{3}{2}-\dfrac{5}{6}\left(\dfrac{1}{2}\right)^2+\sqrt{4}=\dfrac{3}{2}-\dfrac{5}{6}.\dfrac{1}{4}+2=\dfrac{3}{2}-\dfrac{5}{24}+2=\dfrac{36}{24}-\dfrac{5}{24}+\dfrac{48}{24}=\dfrac{79}{24}\)c)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}=1+\left(\dfrac{7}{21}+\dfrac{14}{21}\right)-\dfrac{32}{17}=1+1-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{34}{17}-\dfrac{32}{17}=\dfrac{2}{17}\)d)\(\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)=-8.\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-8.\dfrac{2}{4}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)=-8.\dfrac{2}{4}:\dfrac{13}{12}=-4.\dfrac{12}{13}=\dfrac{-48}{13}\)e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)+28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)=\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right).\left(-\dfrac{5}{3}\right)=\left(\dfrac{120}{7}+\dfrac{196}{7}\right).\left(-\dfrac{5}{3}\right)=\dfrac{316}{7}.\left(-\dfrac{5}{3}\right)=-\dfrac{1580}{21}\)
a, \(\left|2x-3\right|-\dfrac{1}{3}=0\Leftrightarrow\left|2x-3\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
b, tương tự
c, \(\left|2x-1\right|-\left|x+\dfrac{1}{3}\right|=0\Leftrightarrow\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)
TH1 : \(2x-1=x+\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{3}\)
TH2 : \(2x-1=-x-\dfrac{1}{3}\Leftrightarrow3x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{9}\)
d, \(3x-\left|x+15\right|=\dfrac{5}{4}\Leftrightarrow\left|x+15\right|=3x-\dfrac{5}{4}\)ĐK : x >= 5/12
TH1 : \(x+15=3x-\dfrac{5}{4}\Leftrightarrow-2x=-\dfrac{65}{4}\Leftrightarrow x=\dfrac{65}{8}\)( tm )
TH2 : \(x+15=\dfrac{5}{3}-3x\Leftrightarrow4x=-\dfrac{40}{3}\Leftrightarrow x=-\dfrac{10}{3}\)
Ta có : \(12^{30}:36^{15}\) = 1073741824
Mà lại có : \(4^{15}\) = 1073741824
Suy ra thương của \(\dfrac{12^{30}}{36^{15}}\) băng \(4^{15}\)
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