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1) \(4\frac{3}{10}=\frac{43}{10};21\frac{7}{100}=\frac{2107}{100};7\frac{39}{100}=\frac{739}{100};6\frac{123}{1000}=\frac{6123}{1000}\)
2)\(a,5\frac{2}{10}+7\frac{1}{10}=\frac{52}{10}+\frac{71}{10}=\frac{123}{10}\)
\(b,5\frac{6}{7}-3\frac{5}{7}=\frac{41}{7}-\frac{26}{7}=\frac{15}{7}\)
\(c,8\frac{3}{5}x2\frac{6}{7}=\frac{43}{5}x\frac{20}{7}=\frac{172}{7}\)
\(d,1\frac{3}{10}:5\frac{7}{8}=\frac{13}{10}:\frac{47}{8}=\frac{13}{10}x\frac{47}{8}=\frac{611}{80}\)
3) \(7\frac{9}{10}và4\frac{9}{10}\)
Ta có: \(7\frac{9}{10}=\frac{79}{10};4\frac{9}{10}=\frac{49}{10}\)
Suy ra: \(\frac{79}{10}>\frac{49}{10}hay7\frac{9}{10}>4\frac{9}{10}\)
\(6\frac{3}{10}và6\frac{5}{9}\)
Ta có: \(6\frac{3}{10}=\frac{63}{10};6\frac{5}{9}=\frac{59}{9}\)
Suy ra: \(\frac{63}{10}>\frac{59}{9}hay6\frac{3}{10}>6\frac{5}{9}\)
a=1+2-3+4-5+6-7+...+100
a=1+(2-3)+(4-5)+(6-7)+...+(98-99)+100
đặt b=(2-3)+(4-5)+(6-7)+...+(98-99)
b=-1+-1+-1+...+-1(49 số hạng -1)
=>b=-1x49=-49
=>a=1+-49+100
=>a=52
\(2+\frac{3}{10}+\frac{4}{100}\) \(23+\frac{5}{1000}\)
\(=\frac{200}{100}+\frac{30}{100}+\frac{4}{100}\) \(=\frac{23000}{1000}+\frac{5}{1000}\)
\(=\frac{234}{100}=2,34\) \(=\frac{23005}{1000}=23,005\)
\(11+\frac{7}{10}\) \(\frac{6}{10}+\frac{2}{100}\)
\(=\frac{110}{10}+\frac{7}{10}\) \(=\frac{60}{100}+\frac{2}{100}\)
\(=\frac{117}{10}=11,7\) \(=\frac{62}{100}=0,62\)
a) 7/10 :3/10 =7/10*10/3 =70/30 =7/3
b) 9/100:9/10 =9/100*10/9=90/900 =1/10
\(\frac{3}{10}+\frac{7}{100}=\frac{37}{100}\)
\(7+\frac{9}{1000}=\frac{7009}{1000}\)
\(\frac{3}{10}+\frac{7}{100}=\frac{30}{100}+\frac{7}{100}=\frac{37}{100}\)
\(7+\frac{9}{1000}=\frac{7000}{1000}+\frac{9}{1000}=\frac{7009}{1000}\)
a: \(\frac{7}{10}=0,7;\frac{26}{100}=0,26\)
mà 0,7>0,26
nên \(\frac{7}{10}>\frac{26}{100}\)
b: \(\frac{2}{10}=0,2;\frac{46}{100}=0,46\)
mà 0,2<0,46
nên \(\frac{2}{10}<\frac{46}{100}\)
c: \(\frac{3}{10}=0,3;\frac{30}{100}=0,3\)
Do đó: \(\frac{3}{10}=\frac{30}{100}\)
230 + 4 + 3 10 + 7 100 = 230 + 4 + 0 , 3 + 0 , 07 = 230 + 4 , 37 = 234 , 37