biết x+1/x=a
tính x^2+1/(x^2)
x^3+1/(x^3)
x^4+1/(x^4)
x^5+1(x^5) theo a
tích 5 câu đúg đầu tiên
Toán lớp 8
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Câu 1:
a) \(-\dfrac{3}{7}-\left(\dfrac{2}{3}-\dfrac{3}{7}\right)=\dfrac{-3}{7}-\dfrac{2}{3}+\dfrac{3}{7}=\dfrac{-2}{3}\)
Câu 2:
b) \(\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{4}{5}-\dfrac{1}{3}\cdot\dfrac{6}{5}\right)=\dfrac{2}{15}:\left[\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\right]=\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{-2}{5}\right)=\dfrac{2}{15}:\dfrac{-2}{15}=\dfrac{2}{-2}=-1\)
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
1. a
\(\dfrac{8}{5}-\dfrac{5}{6}\cdot\dfrac{3}{4}\)
\(=\dfrac{8}{5}-\dfrac{5\cdot3}{3\cdot2\cdot4}\)
\(=\dfrac{8}{5}-\dfrac{5}{8}=\dfrac{39}{40}\)
1.b
\(=\dfrac{7}{8}+\dfrac{5}{6}\cdot\dfrac{3}{2}\)
\(=\dfrac{7}{8}+\dfrac{5\cdot3}{3\cdot2\cdot2}\)
\(=\dfrac{7}{8}+\dfrac{5}{4}=\dfrac{17}{8}\)
2.a
\(\dfrac{4}{5}+x=\dfrac{11}{10}\)
\(x=\dfrac{11}{10}-\dfrac{4}{5}=\dfrac{3}{10}\)
2.b
\(x-\dfrac{3}{4}=\dfrac{5}{7}\)
\(x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{41}{28}\)
\(E=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2006}\right)\left(1-\frac{1}{2007}\right)\)
\(E=\frac{1}{2}.\frac{2}{3}....\frac{2005}{2006}.\frac{2006}{2007}\)
\(E=\frac{1.2.3.4...2005.2006}{2.3.4.5....2006.2007}\)
\(E=\frac{1}{2007}\)
\(F=1\dfrac{1}{5}\times1\dfrac{1}{6}\times1\dfrac{1}{7}\times\cdot\cdot\cdot\times1\dfrac{1}{2019}\times1\dfrac{1}{2020}\)
\(F=\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}\times\cdot\cdot\cdot\times\dfrac{2020}{2019}\times\dfrac{2021}{2020}\)
\(F=\dfrac{6\times7\times8\times\cdot\cdot\cdot\times2020\times2021}{5\times6\times7\times\cdot\cdot\cdot\times2019\times2020}\)
\(F=\dfrac{2021}{5}\)
\(Huyền\) |
\(f=1^1_5\times1^1_6\times1^1_7\times......\times1^1_{2019}\times1^1_{2022}\)
\(f=\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}\times....\times\dfrac{2020}{2019}\times\dfrac{2021}{2020}\)
\(f=\dfrac{6\times7\times8\times....\times2020\times2021}{5\times6\times7\times.....\times2019\times2020}\)
\(f=\dfrac{2021}{5}\)
\(#Tarus\)