Bài 3:a) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{11}{25}\)\(=\left(\dfrac{5}{15}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)\)\(=\left(\dfrac{5}{15}-\dfrac{20}{15}\right)+1\)\(=-1+1\)\(=0\)b) \(12+8.\left(\dfrac{-1}{2}\right)^3+\left[\left(2018\right)^0:\dfrac{1}{2}\right]^2\)\(=12+8.\dfrac{-1}{8}+\left[1.2\right]^2\)\(=12+-1+2^2\)\(=11+4\)\(=15\)c) \(\left(0,25\right)^{10}.4^{10}+\sqrt{5^2-3^2}\)\(=\left(0,25.4\right)^{10}+\sqrt{25-9}\)\(=1^{10}+\sqrt{16}\)\(=1+4\)\(=5\)Bài 4:a) \(x-\dfrac{7}{5}=0,2\)\(=>x-\dfrac{7}{5}=\dfrac{1}{5}\)\(=>x=\dfrac{1}{5}+\dfrac{7}{5}\)\(=>x=\dfrac{8}{5}\)b) \(\left(x-\dfrac{3}{5}\right):\dfrac{-1}{3}=0,4\)\(=>\left(x-\dfrac{3}{5}\right):\dfrac{-1}{3}=\dfrac{2}{5}\)\(=>x-\dfrac{3}{5}=\dfrac{2}{5}.\dfrac{-1}{3}\)\(=>x-\dfrac{3}{5}=\dfrac{-2}{15}\)\(=>x=\dfrac{-2}{15}+\dfrac{3}{5}=\dfrac{-2}{15}+\dfrac{9}{15}\)\(=>x=\dfrac{7}{15}\)Câu trả lời: Bài 3:a) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{11}{25}\)\(=\left(\dfrac{5}{15}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)\)\(=\left(\dfrac{5}{15}-\dfrac{20}{15}\right)+1\)\(=-1+1\)\(=0\)b) \(12+8.\left(\dfrac{-1}{2}\right)^3+\left[\left(2018\right)^0:\dfrac{1}{2}\right]^2\)\(=12+8.\dfrac{-1}{8}+\left[1.2\right]^2\)\(=12+-1+2^2\)\(=11+4\)\(=15\)c) \(\left(0,25\right)^{10}.4^{10}+\sqrt{5^2-3^2}\)\(=\left(0,25.4\right)^{10}+\sqrt{25-9}\)\(=1^{10}+\sqrt{16}\)\(=1+4\)\(=5\)Bài 4:a) \(x-\dfrac{7}{5}=0,2\)\(=>x-\dfrac{7}{5}=\dfrac{1}{5}\)\(=>x=\dfrac{1}{5}+\dfrac{7}{5}\)\(=>x=\dfrac{8}{5}\)b) \(\left(x-\dfrac{3}{5}\right):\dfrac{-1}{3}=0,4\)\(=>\left(x-\dfrac{3}{5}\right):\dfrac{-1}{3}=\dfrac{2}{5}\)\(=>x-\dfrac{3}{5}=\dfrac{2}{5}.\dfrac{-1}{3}\)\(=>x-\dfrac{3}{5}=\dfrac{-2}{15}\)\(=>x=\dfrac{-2}{15}+\dfrac{3}{5}=\dfrac{-2}{15}+\dfrac{9}{15}\)\(=>x=\dfrac{7}{15}\)