29 + 90 = ........
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\(\frac{36}{47}-\frac{92}{29}-\frac{36}{47}+\frac{90}{29}\)
\(=\left(\frac{36}{47}-\frac{36}{47}\right)+\left(\frac{-92}{29}+\frac{90}{29}\right)\)
\(=0+\frac{-2}{29}\)
\(=\frac{-2}{29}\)
a, \(\dfrac{90}{37}-\dfrac{38}{25}-\dfrac{8}{25}-\dfrac{4}{25}\)
= \(\dfrac{90}{37}\) - \(\dfrac{38+8+4}{25}\)
= \(\dfrac{90}{37}\) - 2
= \(\dfrac{16}{37}\)
\(\dfrac{24}{29}\) + \(\dfrac{32}{41}\) + \(\dfrac{34}{29}\) + \(\dfrac{50}{41}\)
=(\(\dfrac{24}{29}\) + \(\dfrac{34}{29}\)) + (\(\dfrac{32}{41}\) + \(\dfrac{50}{41}\))
= \(\dfrac{58}{29}\) + \(\dfrac{82}{41}\)
= 2 + 2
= 4
\(90+32+29-12-9-70\)
\(=\left(90-70\right)+\left(32-12\right)+\left(29-9\right)\)
\(=20+20+20=60\)
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)
b) Bạn làm tương tự.
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{38}{5}\)
Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)
= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)
B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\)
Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B
Vậy A < B
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + x + 29 = 90
x = 90 - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 29
x= 16
vậy x = 16
\(1+2+3+4+5+6+7+8+9+x+29=90\)
\(\Rightarrow x=90-\left(1+2+3+4+5+6+7+8+9+29\right)\)
\(\Rightarrow x=90-74\)
\(\Rightarrow x=16\)
Vậy x = 16
_Chúc bạn học tốt_
2222222222222222222222222222222*100*200*300*400+1000000000000000000000000*90*18*29-12+222222222222222222222222*11111111 X0=0
Đáp án :
29 + 90 = 119
#Maths
29 + 90 = 119 nha bạn
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