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\(\dfrac{47}{95}\) và \(\dfrac{35}{69}\)
\(\dfrac{47}{95}< \dfrac{1}{2}\) và \(\dfrac{35}{69}>\dfrac{1}{2}\)
Vậy \(\dfrac{47}{95}< \dfrac{35}{69}\)
\(\dfrac{53}{103}\) và \(\dfrac{71}{145}\)
\(\dfrac{53}{103}>\dfrac{1}{2}\) và \(\dfrac{71}{145}< \dfrac{1}{2}\)
Vậy \(\dfrac{53}{103}>\dfrac{71}{145}\)
\(\dfrac{2009}{2010}\) và \(\dfrac{2005}{2006}\)
\(1-\dfrac{2009}{2010}=\dfrac{1}{2010}\) và \(1-\dfrac{2005}{2006}=\dfrac{1}{2006}\)
Vậy \(\dfrac{2009}{2010}>\dfrac{2005}{2006}\)
\(\dfrac{783}{901}\) và \(\dfrac{738}{915}\)
\(\dfrac{738}{915}< \dfrac{783}{915}< \dfrac{783}{901}\)
Vậy \(\dfrac{783}{901}>\dfrac{738}{915}\)
a/
\(\frac{103}{105}=\frac{105-2}{105}=1-\frac{2}{105}=1-\frac{6}{315}\)
\(\frac{205}{208}=\frac{208-3}{208}=1-\frac{3}{208}=1-\frac{6}{416}\)
\(\frac{6}{315}>\frac{6}{418}\Rightarrow\frac{103}{105}< \frac{205}{208}\)
1,
Ta có:
\(\dfrac{73}{75}=1-\dfrac{2}{75}\)
\(\dfrac{77}{79}=1-\dfrac{2}{79}\)
So sánh phân số \(\dfrac{2}{75}\) và \(\dfrac{2}{79}\)
Vì \(75< 79\) nên \(\dfrac{1}{75}>\dfrac{1}{79}\)
Vậy \(1-\dfrac{2}{75}< 1-\dfrac{2}{79}\)
Hay \(\dfrac{73}{75}< \dfrac{77}{79}\)
2,
Vì \(\dfrac{53}{100}>\dfrac{47}{100}>\dfrac{47}{106}\) nên \(\dfrac{53}{100}>\dfrac{47}{106}\)
3,
Ta có:
\(\dfrac{81}{79}=1+\dfrac{2}{79}\)
\(\dfrac{65}{63}=1+\dfrac{2}{63}\)
So sánh phân số \(\dfrac{2}{79}\) và \(\dfrac{2}{63}\)
Vì \(79>63\) nên \(\dfrac{81}{79}< \dfrac{65}{63}\)
Hay \(\Rightarrow1+\dfrac{2}{79}< 1+\dfrac{2}{63}\)
Vậy \(\dfrac{81}{79}< \dfrac{65}{63}\)
4,
\(\dfrac{48}{47}>1>\dfrac{84}{85}\)
Vậy \(\dfrac{48}{47}>\dfrac{84}{85}\)
a) \(\dfrac{81\times49}{7\times8}=\dfrac{81\times7\times7}{7\times8}=\dfrac{81\times7}{8}=\dfrac{567}{8}\)
b) \(\dfrac{105\times69}{3\times35}=\dfrac{3\times35\times69}{3\times35}=69\)
c) \(\dfrac{77\times39}{13\times11}=\dfrac{11\times7\times13\times3}{13\times11}=7\times3=21\)
a) \(\dfrac{81x49}{7x8}=\dfrac{81x7x7}{7x8}=\dfrac{81x7}{8}=\dfrac{567}{8}=70,875\)
b) \(\dfrac{105x69}{3x35}=\dfrac{3x35x69}{3x35}=\dfrac{69}{1}=69\)
c) \(\dfrac{77x39}{13x11}=\dfrac{11x7x3x13}{13x11}=\dfrac{7x3}{1}=\dfrac{21}{1}=21\)
\(=\dfrac{12\times101\times169\times1001}{13\times101\times144\times1001}=\dfrac{13}{12}\)
\(\dfrac{42x45}{5x7}=\dfrac{7x6x5x9}{5x7}=\dfrac{54}{1}=54\)
\(\dfrac{54x56}{7x9}=\dfrac{6x9x7x8}{7x9}=\dfrac{48}{1}=48\)
a: =13/26*9/18*2/3=2/3*1/4=2/12=1/6
b: =44/22*5/15*3/2=2*1/3*3/2=2*1/2=1
( 1 + 1/2 ) . ( 1+ 1/3) . ( 1+ 1/4 ) . ( 1+ 1/5 )
=3/2 . 4/3.5/4.6/5
= 3.4.5.6/2.3.4.5
=6/2 = 3
= 1 x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{5}\) )
= 1 x \(\dfrac{77}{60}\)
= \(\dfrac{77}{60}\)
\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)
\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)
\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)
\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)
\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)
Mẫu số: 8,1: 0,6 x 1875 + 1,5 x 625 x 9
= 9 x 0,9 x 10/6 x 1875 + 0,5 x 3 x 625 x 9
= 9 x 1875 x 1,5 + 9 x 1875 x 0,5
= 9 x 1875 x (1,5 + 0,5)
= 9 x 1875 x 2
= 16875 x 2
Mẫu số: 105 + 205 + 795 + 895
= (105 + 895) + (205 + 795)
= 1000 + 1000
= 1000 x 2
Từ đó, ta có: \(\frac{16875\times2}{1000\times2}=\frac{16875}{1000}=16,875\)
Thật ra là mk biết cách làm rùi nhưng mk muốn các bn làm thử xem cs giống mk làm ko
Ta có:
+) \(\dfrac{103}{105}=\dfrac{105}{105}-\dfrac{2}{105}=1-\dfrac{2}{105}=1-\dfrac{6}{315}\)
+) \(\dfrac{205}{208}=\dfrac{208}{208}-\dfrac{3}{208}=1-\dfrac{3}{208}=1-\dfrac{6}{416}\)
Vì \(\dfrac{6}{315}>\dfrac{6}{416}\) nên \(1-\dfrac{6}{315}< 1-\dfrac{6}{416}\) hay \(\dfrac{103}{105}< \dfrac{205}{208}\)
Ta có:
\(\left\{{}\begin{matrix}1-\dfrac{103}{105}=\dfrac{2}{105}\\1-\dfrac{105}{208}=\dfrac{2}{208}\end{matrix}\right.\)
Vì \(105< 208\Rightarrow\dfrac{2}{105}>\dfrac{2}{208}\Rightarrow\dfrac{103}{105}< \dfrac{203}{208}\)