Question

[Gia Huy To]

Answer 1

[HT.Phong (9A5)]

\(a.\dfrac{2}{3}-\left(-\dfrac{1}{2}-x\right)=-\dfrac{4}{5}\\ \dfrac{2}{3}+\dfrac{1}{2}+x=-\dfrac{4}{5}\\ x=-\dfrac{4}{5}-\dfrac{2}{3}-\dfrac{1}{2}\\ x=-\dfrac{59}{30}\\ b.\left(-x-3\dfrac{1}{4}\right)-\left(1\dfrac{2}{3}-2\dfrac{3}{4}\right)=\dfrac{-5}{6}\\ \left(-x-\dfrac{13}{4}\right)-\left(\dfrac{5}{3}-\dfrac{11}{4}\right)=\dfrac{-5}{6}\\ -x-\dfrac{13}{4}-\dfrac{5}{3}+\dfrac{11}{4}=-\dfrac{5}{6}\\ -x-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{4}{3}\\ c.\dfrac{8}{23}\cdot\dfrac{46}{24}-\dfrac{1}{2}x=\dfrac{1}{3}\\ \dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ d.\dfrac{x-1}{16}=\dfrac{3}{x+1}\\ \left(x-1\right)\left(x+1\right)=3\cdot16=48\\ x^2-1=48\\ x^2=49\\ x^2=7^2\\ x=\pm7\)

Answer 2

[HT.Phong (9A5)]

\(e.\left(1,2\right)^3x^2=\left(1,2\right)^5\\ x^2=\dfrac{\left(1,2\right)^5}{\left(1,2\right)^3}\\ x^2=\left(1,2\right)^2\\ x=\pm1,2\\ f.\left(\dfrac{2}{3}x-\dfrac{1}{4}\right)^2=4\\ \left(\dfrac{2}{3}x-\dfrac{1}{4}\right)^2=2^2\\TH1:\dfrac{2}{3}x-\dfrac{1}{4}=2\\ \dfrac{2}{3}x=2+\dfrac{1}{4}=\dfrac{9}{4}\\ x=\dfrac{9}{4}:\dfrac{2}{3}=\dfrac{27}{8}\\ TH2:\dfrac{2}{3}x-\dfrac{1}{4}=-2\\ \dfrac{2}{3}x=-2+\dfrac{1}{4}=-\dfrac{7}{4}\\ x=\dfrac{-7}{4}:\dfrac{2}{3}=-\dfrac{21}{8}\\ g.\left(\dfrac{1}{6}x-3\right)^2=\dfrac{4}{9}\\ \left(\dfrac{1}{6}x-3\right)^2=\left(\dfrac{2}{3}\right)^2\\ TH1:\dfrac{1}{6}x-3=\dfrac{2}{3}\\ \dfrac{1}{6}x=\dfrac{2}{3}+3=\dfrac{11}{3}\\ x=\dfrac{11}{3}:\dfrac{1}{6}=22\\ TH2:\dfrac{1}{6}x-3=-\dfrac{2}{3}\\ \dfrac{1}{6}x=-\dfrac{2}{3}+3=\dfrac{7}{3}\\ x=\dfrac{7}{3}:\dfrac{1}{6}=14\)