Bài `1 a,`
`A = 5/15 + 14/25 - 12/9 + 2/7 + 11/25`
`A = [(5/15 - 12/9) + (14/25 + 11/25)] + 2/7`
`A = [(-1) + (1)] + 2/7`
`A = 0 + 2/7`
`A = 2/7`
a: \(A=\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}\)
\(=-1+1+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(B=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)
\(A=\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\)
\(A=\left[\left(\dfrac{5}{15}-\dfrac{12}{9}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)\right]+\dfrac{2}{7}\)
\(A=\left[\left(\dfrac{15}{45}-\dfrac{60}{45}\right)+\dfrac{25}{25}\right]+\dfrac{2}{7}\)
\(A=\left[\dfrac{-45}{45}+1\right]+\dfrac{2}{7}\)
\(A=\left[\left(-1\right)+1\right]+\dfrac{2}{7}\)
\(A=0+\dfrac{2}{7}\)
\(A=\dfrac{2}{7}\)
\(B=\dfrac{2.8^4.27^2+4.6^9}{2^7.6^7+2^7.40.9^4}\)
\(B=\dfrac{2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9}{2^7.\left(2.3\right)^7+2^7.2^3.5.\left(3^2\right)^4}\)
\(B=\dfrac{2.2^{12}.3^6+2^2.2^9.3^9}{2^7.2^7.3^7+2^7.2^3.5.3^8}\)
\(B=\dfrac{2^{13}.3^6+2^{11}.3^9}{2^{14}.3^7+2^{10}.5.3^8}\)
\(B=\dfrac{\left(2^2+3^3\right).2^{11}.3^6}{\left(2^4+5.3\right).2^{10}.3^7}\)
\(B=\dfrac{31.2^{11}.3^6}{31.2^{10}.3^7}\)
\(B=\dfrac{2}{3}\)
