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a: 2003.2004-1/2003.2004 < 2004.2005-1/2004.2005
b: 149/157 < 449/457
c: 1999.2000/1999.2000+1 < 2000.2001/2000.2001+1
Bài 1 :
\(a)\frac{-17}{30}-\frac{11}{-15}+\left(-\frac{7}{12}\right)\)
\(=\frac{1}{6}+\left(-\frac{7}{12}\right)\)
\(=-\frac{5}{12}\)
\(b)-\frac{5}{9}+\frac{5}{9}:\left(1\frac{2}{3}-2\frac{1}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{9}:\left(\frac{5}{3}-\frac{25}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{9}:\left(-\frac{5}{12}\right)\)
\(=-\frac{5}{9}+\left(-\frac{2}{3}\right)\)
\(=-\frac{1}{9}\)
\(c)-\frac{7}{25}\times\frac{11}{13}+\left(-\frac{7}{25}\right)\times\frac{2}{13}-\frac{18}{25}\)
\(=-\frac{77}{325}+\left(-\frac{14}{325}\right)-\frac{18}{25}\)
\(=-\frac{7}{25}-\frac{18}{25}\)
\(=-1\)
ta sẽ phải dùng pp phần bù 1-1999.2000/1999.2001=1/2001 1-2000.2001/1999.2002=1/2002 ta thấy : cùng tử nhưng mẫu số của phân số nào bé hơn thì phân số đó lớn hơn =>1.2001<1/2002 thì 1/2001 >1/2002 =>1999.2000/1999.2001>2000.2001/2000.2001=>1999.2000/1999.2000+1>2000.2001/2000.2001+1 vậy 1999.2000/1999.2000+1>2000.2001/2000.2001+1 / là phân số
https://www.youtube.com/channel/UC5odkiOvzz9Rvu3HUYlL2IQ?view_as=subscriber
\(A=\frac{1999.2000+1-1}{1999.2000+1}=1-\frac{1}{1999.2000+1}\)
\(B=1-\frac{1}{2000.2001+1}\)
1999.2000+1 < 2000.2001+1
nên 1/1999.2000+1 > 1/2000.2001+1
nên 1 - 1/1999.2000+1 < 1 - 1/2000.2001+1
Vậy A < B
Bài 1:
\(\frac{3}{5}+\frac{4}{15}=\frac{9}{15}+\frac{4}{15}=\frac{13}{15}\)
\(\frac{5}{6}:\frac{-7}{12}=\frac{5}{6}.\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
\(\frac{-21}{24}:\frac{-14}{8}=\frac{-21}{24}.\frac{-8}{14}=\frac{168}{336}=\frac{1}{2}\)
\(\frac{4}{5}:\frac{-8}{15}=\frac{4}{5}.\frac{-15}{8}=\frac{-60}{40}=\frac{-3}{2}\)
\(\frac{5}{12}-\frac{-7}{6}=\frac{5}{12}+\frac{7}{6}=\frac{5}{12}+\frac{14}{12}=\frac{19}{12}\)
\(\frac{-15}{16}.\frac{8}{25}=\frac{-120}{400}=\frac{-3}{10}\)
Bài 2 :
\(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
\(=\frac{34}{5}-\left(\frac{5}{3}+\frac{19}{5}\right)\)
\(=\frac{34}{5}-\frac{5}{3}-\frac{19}{5}\)
\(=\left(\frac{34}{5}-\frac{19}{5}\right)-\frac{5}{3}\)
\(=3-\frac{5}{3}\)
\(=\frac{4}{3}\)
\(6\frac{5}{7}-\left(1\frac{2}{3}+2\frac{5}{7}\right)\)
\(=\frac{47}{7}-\left(\frac{5}{3}+\frac{19}{7}\right)\)
\(=\frac{47}{7}-\frac{5}{3}-\frac{19}{7}\)
\(=\left(\frac{47}{7}-\frac{19}{7}\right)-\frac{5}{3}\)
\(=4-\frac{5}{3}\)
\(=\frac{7}{3}\)
\(\frac{4}{19}.\frac{-3}{7}+\frac{-3}{7}.\frac{15}{19}+\frac{5}{7}\)
\(=\left(\frac{4}{19}+\frac{15}{19}\right).\frac{-3}{7}+\frac{5}{7}\)
\(=1.\frac{-3}{7}+\frac{5}{7}\)
\(=\frac{-3}{7}+\frac{5}{7}\)
\(=\frac{2}{7}\)
\(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
\(=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
|x+3|=|-9|
TH1: x+3=9 => x=9-3 TH2: x+3=-9=> x=-9 -3
x=6 x=-12