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b) \(\frac{5}{9}\)và \(\frac{5}{8}\) :Quy đồng mẫu số : \(\frac{5}{9}\) = \(\frac{5.8}{9.8}\) = \(\frac{40}{72}\) ; \(\frac{5}{8}\) = \(\frac{5.9}{8.9}\) = \(\frac{45}{72}\)
Vì \(\frac{40}{72}\) < \(\frac{45}{72}\) nên \(\frac{5}{9}\) < \(\frac{5}{8}\)
c)\(\frac{8}{7}\) và \(\frac{7}{8}\) :Quy đồng mẫu số: \(\frac{8}{7}\) = \(\frac{8.8}{7.8}\) = \(\frac{64}{56}\) ; \(\frac{7}{8}\) = \(\frac{7.7}{8.7}\) =\(\frac{49}{56}\)
Vì \(\frac{64}{56}\) > \(\frac{49}{56}\) nên \(\frac{8}{7}\) > \(\frac{7}{8}\)
bạn an đông à cái câu A của bạn sai một chút.
CHÚC BẠN HỌC TỐT !
a)\(\frac{3}{7}\) và\(\frac{2}{8}\) :Quy đồng mẫu số : \(\frac{3}{7}\) = \(\frac{3.8}{7.8}\) = \(\frac{24}{56}\) ; \(\frac{2}{8}\) = \(\frac{2.7}{8.7}\) = \(\frac{14}{56}\)
Vì \(\frac{24}{56}\) > \(\frac{14}{56}\) nên \(\frac{3}{7}\) > \(\frac{2}{8}\)
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)\(\frac{77}{24}\)=\(3\frac{5}{24}\)
b)=10
c)\(\frac{24}{35}\)
1)
a) \(x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=5\)
\(x+\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}=5\)
\(x+\frac{127}{128}=5\)
\(x=5-\frac{127}{128}=\frac{513}{128}\)
b) \(x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}=3\)
\(x+\frac{729}{2187}+\frac{243}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}=3\)
\(x+\frac{2186}{2187}=3\)
\(x=3-\frac{2186}{2187}=\frac{4375}{2187}\)
2)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
b) \(5\frac{1}{2}+3\frac{5}{6}+\frac{2}{3}\)
\(=\left(5+3\right)+\left(\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\right)\)
\(=8+\left(\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\right)\)
\(=8+2=10\)
c) \(7\frac{7}{8}+1\frac{4}{6}+3\frac{3}{5}\)
\(=\left(7+1+3\right)+\left(\frac{7}{8}+\frac{2}{3}+\frac{3}{5}\right)\)
\(=11+\left(\frac{105}{120}+\frac{80}{120}+\frac{72}{120}\right)\)
\(=11+\frac{257}{120}=\frac{1577}{120}\)
3) Gọi số đó là x. Theo đề ta có :
\(\frac{16-x}{21+x}=\frac{5}{7}\)
\(7\left(16-x\right)=5\left(21+x\right)\)
\(112-7x=105+5x\)
\(112-105=7x-5x\)
\(7=2x\)
\(x=\frac{7}{2}=3,5\) ( vô lí )
Vậy không có số tự nhiên để thõa mãn điều kiện trên.
tl :
\(\frac{5}{14}\)= \(\frac{5x21}{14x21}\)= \(\frac{105}{294}\)
\(\frac{4}{21}\)= \(\frac{4x14}{21x14}\)= \(\frac{56}{294}\)
a,(11/15+4/15)+(5/7+2/7)
=1+1
=2
b,5/9x(1/2+6/4)
=5/9x2
=10/9
c,1/2:(7/8+9/8)
=1/2:2
=1
d,(17/10-7/10)+1/2
=1+1/2
=3/2