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Câu hỏi của ๖Eric Gaming - Toán lớp 4 - Học toán với OnlineMath
Bài 2:
\(=\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)=2+\dfrac{1}{2}=\dfrac{5}{2}\)
Bài 1:
150/120+15/27=5/4+5/9=45/36+20/36=65/36
42/49+56/84=6/7+2/3=9/21+14/21=23/21
= (27/12 + 25/36)+(17/6 + 15/6)= 81/36 + 25/36 + 17/6 + 15/6 = 106/36 + 32/6 = 192/36 + 106/36 = 298/36 = 149/18
\(\dfrac{5}{7}\)>\(\dfrac{6}{21}\)
\(\dfrac{18}{24}\)>\(\dfrac{15}{25}\)
\(\dfrac{12}{13}\)=\(\dfrac{1212}{1313}\)
1010 + 1111 + 1212 + 1313 + ... + 9898 + 9999
= 101 x 10 + 101 x 11 + 101 x 12 + .... 101 x 98 + 101 x 99
= 101 x ( 10 + 11 + 12 + ... + 98 + 99 )
= 101 x 109 x 45
= 495405
\(1235\text{×}6789\text{×}\left(630-315\text{×}2\right)\)
\(=1235\text{×}6789\text{×}\left(630-630\right)\)
\(=1235\text{×}6789\text{×}0=0\)
\(\left(100+67\right)\text{×}67+\left(200-38\right)\text{×}33\)
\(=167\text{×}67+167\text{×}33\)
\(=167\text{×}\left(67+33\right)\)
\(=167\text{×}100=16700\)
\(72+36\text{×}2+24\text{×}3+18\text{×}4+12\text{×}2+168\)
\(=72+72+72+72+24+168\)
\(=\left(72\text{×}4\right)+24+168=288+24+168=480\)
1. \(\frac{12}{18}\)= \(\frac{2}{3}\)
\(\frac{22}{33}\)= \(\frac{2}{3}\)
\(\frac{1010}{1313}\)= \(\frac{10}{13}\)
\(\frac{27}{36}\)= \(\frac{3}{4}\)
\(\frac{202}{303}\)= \(\frac{2}{3}\)
\(\frac{1111}{1212}\)= \(\frac{11}{12}\)
2. a) 12 x \(\frac{1313}{13}\)x 1212
= 12 x \(\frac{13}{13}\)x 1212
1. \(\frac{12}{18}=\frac{2}{3}\): \(\frac{22}{33}=\frac{2}{3}\); \(\frac{1010}{1313}=\frac{10}{13}\); \(\frac{27}{36}=\frac{3}{4}\) ; \(\frac{202}{303}=\frac{2}{3}\);\(\frac{1111}{1212}=\frac{11}{12}\)
2 a)\(\frac{12X1313}{13X1212}=\frac{12X13X101}{13X12X101}=1\);b) \(\frac{1111X121212}{1212X111111}=\frac{11X101X12X10101}{12X101X11X10101}=1\)