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\(\frac{3457}{3452}=\frac{3452+5}{3452}=1+\frac{5}{3452},\frac{6789}{6784}=\frac{6784+5}{6789}=1+\frac{5}{6789}\)
Ta có: \(3452< 6789\Leftrightarrow\frac{1}{3452}>\frac{1}{6789}\Leftrightarrow\frac{5}{3452}>\frac{5}{6789}\)
suy ra \(\frac{3457}{3452}>\frac{6789}{6784}\).
3457/3452=1+5/3452
6789/6784=1+5/6784
mà 3452>6784
nên 3457/3452<6789/6784
a:
13/17=1-4/17
8/12=1-4/12
mà 4/17<4/12
nên 13/17>8/12=12/18
b: 16/51<17/51=1/3=30/90<31/90
a)
\(\dfrac{5}{9}< \dfrac{9}{9}\)
\(\dfrac{8}{7}>\dfrac{7}{7}\)
\(\dfrac{9}{9}=1\)
\(\dfrac{18}{4}>\dfrac{3}{4}\)
b)
\(\dfrac{2}{5},\dfrac{3}{5},\dfrac{8}{5}\)
\(\dfrac{5}{2}=\dfrac{15}{6},\dfrac{1}{6},1=\dfrac{6}{6}\rightarrow\dfrac{1}{6},\dfrac{6}{6},\dfrac{15}{6}\)
a) \(\dfrac{3}{4}=\dfrac{3\times4}{4\times4}=\dfrac{12}{16}\)
b) \(\dfrac{1}{3}=\dfrac{1\times3}{3\times3}=\dfrac{3}{9}\)
c) \(\dfrac{5}{6}=\dfrac{5\times3}{6\times3}=\dfrac{15}{18}\)
2003 / 2001 = 1 + 2/2001
1999/1997 = 1 + 2/1997
vì 2/ 2001 < 2/1997
nên 1 + 2/2001 < 1 + 2/1997
hay 2003 < 1999/1997
b, = 5/9 x 1/4 + 4/9 x 1/4
= 1/4 x ( 5/9 + 4/9 )
= 1/4 x 1
= 1/4
* Ý a mk k nhớ cách làm ^^, xl *
\(b,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}\)
\(=\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}\)
\(=\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{1}{4}\times\dfrac{9}{9}=\dfrac{1}{4}\times1=\dfrac{1}{4}\)
a) \(\dfrac{2}{5}=\dfrac{4}{10}\)
\(\dfrac{4}{10}>\dfrac{3}{10}\)
b) \(\dfrac{5}{6}=\dfrac{10}{12}\)
\(\dfrac{7}{12}< \dfrac{10}{12}\)
c) \(\dfrac{1}{2}=\dfrac{2}{4}\)
\(\dfrac{3}{4}< \dfrac{2}{4}\)
d) \(\dfrac{8}{3}=\dfrac{56}{21}\)
\(\dfrac{56}{21}>\dfrac{11}{21}\)
3457/3452 > 6789/6784
Ta có:
3457/3452=1+(5/3452)
6789/6784=1+(5/6784)
Vì 1=1 nên ta so sánh 5/3452 và 5/6784
Vì 5=5, 3452<6784
Nên 5/3452>5/6784
Do đó:1+5/3452>1+5/6784
hay 3457/3452>6789/6784
Đây nha