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= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)
= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)
= \(x^8.\frac{1}{10}.1.1.1.1\)
= \(x^8.\frac{1}{10}\)
Mk ko pik co dung ko nua
1/3xD=1/(2x4)+1/(4x6)+...+1/(98x100)
2/3xD=2/(2x4)+2/(4x6)+...+1/(98x100)
2/3xD= 1/2-1/4+1/4-1/6+...+1/98-1/100
2/3xD=1/2-1/100
2/3xD=49/100
D=147/200
a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2
=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)
=> \(-2x=\frac{-85}{4}\)
=> \(x=\frac{-85}{4}:\left(-2\right)\)
=> \(x=\frac{85}{8}\)
b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)
=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)
=> \(\frac{-2}{3}x=\frac{15}{29}\)
=> x = \(\frac{15}{29}:\frac{-2}{3}\)
=> x = \(\frac{-45}{58}\)
\(\frac{3}{2}+\frac{3}{4}+\frac{3}{6}+\frac{3}{8}+\frac{3}{10}+\frac{3}{12}\)
\(=\frac{18}{12}+\frac{9}{12}+\frac{6}{12}+\frac{3}{12}+\frac{3}{10}\)
\(=3+\frac{3}{10}\)
\(=\frac{30}{10}+\frac{3}{10}\)
\(=\frac{33}{10}\)
Ta thấy tất cả các phân số đều có mẫu số chung bằng 120
=> \(\frac{3}{2}\)+ \(\frac{3}{4}\)+ \(\frac{3}{6}\)+ \(\frac{3}{8}\)+ \(\frac{3}{10}\)+ \(\frac{3}{12}\)
= \(\frac{18}{120}\)+ \(\frac{90}{120}\)+ \(\frac{60}{120}\)+ \(\frac{45}{120}\)+ \(\frac{36}{120}\)+ \(\frac{30}{120}\)( cộng các ps lại với nhau )
= \(\frac{279}{120}\)
nhớ ủng hộ mik nghen