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14 tháng 3 2017

dễ mà

\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)thay x = 1;x=3 vào ra kq thui

16 tháng 3 2017

thanks nha bạn

31 tháng 12 2022

\(C=\dfrac{4}{9}\times\dfrac{13}{17}+\dfrac{4}{17}\times\dfrac{4}{9}+\dfrac{2}{9}\\ =\dfrac{4}{9}\times\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{2}{9}\\ =\dfrac{4}{9}\times1+\dfrac{2}{9}\\ =\dfrac{4}{9}+\dfrac{2}{9}\\ =\dfrac{6}{9}=\dfrac{2}{3}\)

\(D=\dfrac{8}{19}\times\dfrac{5}{11}+\dfrac{7}{11}\times\dfrac{8}{19}+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{8}{19}\times\left(\dfrac{5}{11}+\dfrac{7}{11}\right)+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{8}{19}\times\dfrac{12}{11}+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{12}{11}\times\left(\dfrac{8}{19}+\dfrac{11}{19}\right)\\ =\dfrac{12}{11}\times19\\ =\dfrac{12}{11}\)

31 tháng 12 2022

\(C=\dfrac{4}{9}\cdot\dfrac{13}{17}+\dfrac{4}{17}\cdot\dfrac{4}{9}+\dfrac{2}{9}\)

\(C=\dfrac{4}{9}\cdot\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{2}{9}\)

\(C=\dfrac{4}{9}\cdot\dfrac{13+4}{17}+\dfrac{2}{9}\)

\(C=\dfrac{4}{9}\cdot\dfrac{17}{17}+\dfrac{9}{2}\)

\(C=\dfrac{4}{9}\cdot1+\dfrac{2}{9}\)

\(C=\dfrac{4}{9}+\dfrac{2}{9}\)

\(C=\dfrac{4+2}{9}\)

\(C=\dfrac{6}{9}\)

\(C=\dfrac{2}{3}\)

 

\(D=\dfrac{8}{19}\cdot\dfrac{5}{11}+\dfrac{7}{11}\cdot\dfrac{8}{19}+\dfrac{12}{11}\cdot\dfrac{11}{19}\)

\(D=\dfrac{8}{19}\cdot\left(\dfrac{5}{11}+\dfrac{7}{11}\right)+\dfrac{12}{11}\cdot\dfrac{11}{19}\)

\(D=\dfrac{8}{19}\cdot\dfrac{12}{11}+\dfrac{12}{11}\cdot\dfrac{11}{19}\)

\(D=\dfrac{12}{11}\cdot\left(\dfrac{8}{19}+\dfrac{11}{19}\right)\)

\(D=\dfrac{12}{11}\cdot\dfrac{19}{19}\)

\(D=\dfrac{12}{11}\cdot1\)

\(D=\dfrac{12}{11}\)

28 tháng 3 2022

Hơi choáng

28 tháng 3 2022

Bạn đăng 2 câu 1 lần đc ko

27 tháng 2 2022

\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)

\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)

\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)

\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)

\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)

27 tháng 2 2022

Viết lại phần d) đc 0 ạ=((

giúp mk vs mn ơi , mai cô giáo ktra mk r

a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)

b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)

c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)

c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)

a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)

b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)

6 tháng 2 2022

nhờ bạn có thể giải chi tiết cho mình câu 1b đc ko

23 tháng 4 2023

1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{19}{18}+\dfrac{5}{18}\)

\(=\dfrac{24}{18}\)

\(=\dfrac{4}{3}\)

2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\dfrac{1}{15}+\dfrac{7}{15}\)

\(=\dfrac{8}{15}\)

3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)

\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

\(=\dfrac{7}{30}\)

4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}.-\dfrac{7}{11}\)

\(=-\dfrac{35}{77}\)

\(=-\dfrac{5}{11}\)

Chọn B

NV
18 tháng 3 2023

\(=\left(log_{a^{-1}}a^2\right)^2+\dfrac{1}{2}.\dfrac{1}{2}log_aa\)

\(=\left(-1.2.log_aa\right)^2+\dfrac{1}{4}=4+\dfrac{1}{4}=\dfrac{17}{4}\)