=\(\frac{587}{140}\)

t.i.c.k nha

#)Trả lời : 

\(\frac{81}{10}\)\(\frac{81}{10}\)

Bn khác giải ra nha, kết quả đó, dài lắm, lười :V

       #~Will~be~Pens~#

\(\frac{15}{37}\)x \(\left(\frac{38}{41}-\frac{74}{45}\right)-\frac{38}{41}\)x \(\left(\frac{15}{37}+\frac{82}{76}\right)\)

\(=\frac{15}{37}\)x \(\frac{38}{41}-\frac{15}{37}\)x \(\frac{74}{45}-\frac{38}{41}\)x \(\frac{15}{37}-\frac{38}{41}\)x \(\frac{82}{76}\)

\(=\frac{15}{37}\)x  \(\frac{38}{41}-\frac{2}{3}-\frac{38}{41}\)x  \(\frac{15}{37}-1\)

\(=\frac{15}{37}\)x  \(\frac{38}{41}\)\(-\frac{38}{41}\)x   \(\frac{15}{37}\)\(-\frac{2}{3}-1\)

\(=0-\frac{2}{3}-1=\frac{-5}{3}\)

đừng chơi với ngu vip

7/41 + 5/9 + -7/41 + 16 / 9

\(= \frac{7}{41} - \frac{7}{41} + \frac{5}{9} + \frac{16}{9}\)

\(\frac{7}{41} - \frac{7}{41} = 0\)

\(0 + \frac{5}{9} + \frac{16}{9}\)

\(\frac{5}{9} + \frac{16}{9} = \frac{21}{9}\)

\(\frac{21}{9}=\frac{7}{3}\)

**Trả lời:
\(\frac{7}{41}+\frac59+\frac{-7}{41}+\frac{16}{9}\)
\(=\left(\frac{7}{41}+\frac{-7}{41}\right)+\left(\frac59+\frac{16}{9}\right)\)
\(=0+\frac{21}{9}\)
\(=\frac{21}{9}.\)

-42/32.(15/8-16/41)+15/8.(41/32.8/3)

=-42/32.487/328+15/8.41/12

=-10227/5248+205/32

=23393/5248

a) \(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)

\(=\frac{31}{23}-\frac{15}{23}\)

\(=\frac{16}{23}\)

b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)

\(=\frac{1}{3}-1+1\)

\(=\frac{1}{3}\)

c) \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{52}-\frac{3}{11}\right)\)

\(=\frac{38}{45}-\frac{8}{45}+\frac{17}{52}+\frac{3}{11}\)

\(=\frac{30}{45}+\frac{17}{52}+\frac{3}{11}\)

\(=\frac{2}{3}+\frac{17}{52}+\frac{3}{11}\)

\(=\frac{104+51}{156}+\frac{3}{11}\)

\(=\frac{155}{156}+\frac{3}{11}\)

\(=\frac{156}{156}-\frac{1}{156}+\frac{3}{11}\)

\(=1-\frac{1}{156}+\frac{3}{11}\)

\(=1-\left(\frac{11-468}{1716}\right)\)

\(=1-\frac{-457}{1716}\)

\(=1+\frac{457}{1716}\)

\(=\frac{2173}{1716}\)

a)31/23-(7/32+8/23)=31/23-7/32-8/23=(31/23-8/23)-7/32=1-7/32=25/32

a) 7/15-(2/15-12/18)

=7/15-2/15+2/3

=5/15+2/3

=1/3+2/3

=1

b)(7/41- 4/9) - (3/19 + 7/41) + (4/9 - 16/19)

=7/41 - 4/9 - 3/19 - 7/41 + 4/9 + 16/19

=(7/41 - 7/41) - (4/9 - 4/9) - (3/19 + 16/19)

= -1

b) 7/41 - 4/9 - 3/19 - 7/41 + 4/9 + 16/19

= (7/41 - 7/41 ) - (4/9 - 4/9 ) + ( 3/19 + 16/19 )

= 0 - 0 + 1

= 1

Ta có : 

\(\frac{7}{12}\)= \(\frac{4}{12}\)+ \(\frac{3}{12}\)= \(\frac{1}{3}\)+ \(\frac{1}{4}\)= \(\frac{20}{60}\)+ \(\frac{20}{80}\)

\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ .... + \(\frac{1}{79}\)+ \(\frac{1}{80}\)= (\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ ....+\(\frac{1}{60}\)) + ( \(\frac{1}{61}\)+ \(\frac{1}{62}\)+...+\(\frac{1}{79}\)+\(\frac{1}{80}\))

Do \(\frac{1}{41}\)>\(\frac{1}{42}\)>....>\(\frac{1}{60}\)

=> ( \(\frac{1}{41}\)+ \(\frac{1}{42}\)+...+\(\frac{1}{60}\)) > \(\frac{1}{60}\)+...+\(\frac{1}{60}\)= \(\frac{20}{60}\)

Vậy : \(\frac{1}{61}\)> \(\frac{1}{62}\)>....>\(\frac{1}{79}\)>\(\frac{1}{80}\)

=> ( \(\frac{1}{61}\)+\(\frac{1}{62}\)+...+\(\frac{1}{79}\)+ \(\frac{1}{80}\)) > \(\frac{1}{80}\)+...+ \(\frac{1}{80}\)= \(\frac{20}{80}\)

Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+\(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)

Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+ \(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)= \(\frac{7}{12}\)

=> ĐPCM