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\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right)\)
\(=\frac{3}{5}+\frac{3}{11}+\frac{3}{7}+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}-\frac{23}{44}\)
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\frac{35}{35}+\left(\frac{-44}{44}\right)+\frac{2}{97}=1+\left(-1\right)+\frac{2}{97}=\frac{2}{97}\)
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=-1+1+\frac{2}{97}\)
\(=\frac{2}{97}\)
\(=\left[\frac{3}{5}-\frac{1}{35}-\left(\frac{-3}{7}\right)\right]+\left[\frac{3}{11}-\frac{3}{4}+\left(\frac{-23}{44}\right)\right]\)
\(=\left[\frac{21}{35}-\frac{1}{35}+\frac{15}{35}\right]+\left[\frac{12}{44}-\frac{33}{44}+\left(\frac{-23}{44}\right)\right]\)
\(=\left[\frac{20}{35}+\frac{15}{35}\right]+\left[\frac{-21}{44}+\left(\frac{-23}{44}\right)\right]\)
\(=1+\left(-1\right)\)
\(=0\)
\(=\dfrac{3}{5}+\dfrac{3}{11}+\dfrac{3}{7}-\dfrac{1}{35}-\dfrac{3}{4}+\dfrac{-23}{44}\)
\(=\left(\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{1}{35}\right)+\left(\dfrac{3}{11}-\dfrac{3}{4}-\dfrac{23}{44}\right)\)
\(=\dfrac{21+15-1}{35}+\dfrac{12-33-23}{44}\)
\(=1-1=0\)
b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)
\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)
\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)
\(=\frac{22}{29}.\)
Chúc bạn học tốt!
câu b nha
B= 1/100 - (1/2.1 + 1/3.2 + ... + 1/98.97 + 1/99.98 + 1/100.99)
B=1/100 - (1 - 1/2 + 1/2 - 1/3 + 1/3 - ... - 1/99 + 1/99 - 1/100)
B=1/100-(1-1/100)
B=1/100-99/100
B= - 98/100
B= - 49/50
đ ú g nha
1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
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\(=\frac{-305}{679}\)