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Kết quả :
\(=\frac{37}{53}x\frac{23}{48}x\frac{53}{37}x\frac{24}{23}\)
\(=\left(\frac{37}{53}x\frac{53}{37}\right)x\left(\frac{23}{48}x\frac{24}{23}\right)\)
\(=1x\frac{24}{48}\)
\(=1x\frac{1}{2}\)
\(=\frac{1}{2}\)
5/11+1/2+2/5+6/11+3/4+16/25+5/16
=1441/400
19/4+37/100+1/8+132/25+5/2+37/12
=1933/120
=54.275+275+825.15
=275.55+825.15
=275.55+55.15.15
=55.(275+225)
=55.500
=27500
\(\frac{1997x1996-1}{1995x1997+1996}\)
\(=\frac{1995x1997+1997-1}{1995x1997+1996}\)
\(=\frac{1995x1997+1996}{1995x1997+1996}\)
\(=1\)
\(\frac{1996x1995-996}{1000+1996x1994}\)
\(=\frac{1996x1994+1996-996}{1000+1996x1994}\)
\(=\frac{1996x1994+1000}{1000+1996x1994}\)
\(=1\)
1996 x 1995 - 1996 - 1996 x 1994 =
1996 x 1995 - 1996 x 1995 = 0
k cho mk nha!!
a) Ta có:
\(\frac{23}{3}=\frac{21+2}{3}=\frac{21}{3}+\frac{2}{3}=7+\frac{2}{3}\)
\(\frac{15}{2}=\frac{14+1}{2}=\frac{14}{2}+\frac{1}{2}=7+\frac{1}{2}\)
Lại có:\(1-\frac{2}{3}=\frac{1}{3}< 1-\frac{1}{2}=\frac{1}{2}\Rightarrow\frac{2}{3}>\frac{1}{2}\)
\(\Rightarrow\frac{23}{3}>\frac{15}{2}\)
a, 23/3 và 15/2
Ta có :
23/3 = 23x2/3x2 = 46/6. 15/2 = 15x3/2x3 = 45/6.
Vậy 46/6 > 45/6 nên 23/3 > 15/2.
b, 4/17 và 2/9.
Ta có :
4/17 = 4x9/17x9 = 36/153. 2/9 = 2x17/9x17 = 34/153.
Vậy 36/153 > 34/153 nên 4/17 > 2/9.
=\(\frac{37}{53}+\frac{23}{48}+\frac{53}{37}+\frac{24}{23}\)
=\(\left(\frac{37}{53}+\frac{53}{37}\right)+\left(\frac{23}{24.2}+\frac{24}{23}\right)\)
=\(1+\left(\frac{23}{24}-\frac{23}{2}+\frac{23}{24}\right)\)
=\(1+\frac{23}{2}=\frac{25}{2}\)