So sánh 2134/2135 với 1439/1440
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\(3^{135}+2^{135}+3^{133}+2^{134}\)
\(=\left(3^{135}+3^{133}\right)+\left(2^{135}+2^{134}\right)\)
\(=3^{133}\cdot\left(3^2+1\right)+2^{133}\cdot\left(4+1\right)\)
\(=3^{133}\cdot10+2^{133}\cdot5\)
\(=5\cdot2\cdot\left(3^{133}+2^{132}\right)\)
\(=10\cdot\left(3^{133}+2^{132}\right)\)
a)
\(a=2136\times2136=2136\times\left(2134+2\right)=2136\times2134+2136\times2\)
\(b=2134\times2138=2134\times\left(2136+2\right)=2134\times2136+2134\times2\)
Vì \(2136\times2>2134\times2\)
=> \(a>b\)
b)
=> \(a=b\)\(b=123\times246+77-=123\times\left(245+1\right)+77=123\times245+123+77=123\times245+200\)\(a=245\times124-45=245\times\left(123+1\right)-45=245\times123+245-45=245\times123+200\)
A = 2136 x 2136 = (2134 + 2) x 2136
= 2134 x 2136 + 2 x 2136
B = 2134 x 2138
B = 2134 x ( 2136 + 2 )
B = 2134 x 2136 + 2 x 2134.
Vì 2134 x 2136 = 2134 x 2136 và 2 x 2136 > 2 x 2134 => 2136 x 2136 > 2134 x 2138 => A > B
\(a)\dfrac{-11}{12}và\dfrac{17}{-18}\) \(\Leftrightarrow\dfrac{-11}{12}và\dfrac{-17}{18}\) \(\Leftrightarrow\dfrac{-33}{36}và\dfrac{-34}{36}\)
Ta thấy rằng : \(-33>-34\Rightarrow\dfrac{-33}{36}>\dfrac{-34}{36}\)
Hay : \(\dfrac{-11}{12}>\dfrac{17}{-18}\)
\(b)\dfrac{-14}{-21}và\dfrac{-60}{-72}\)
Ta có : \(\dfrac{-14}{-21}\text{=}\dfrac{-14:-7}{-21:-7}\text{=}\dfrac{2}{3}\text{=}\dfrac{4}{6}\)
\(\dfrac{-60}{-72}\text{=}\dfrac{-60:-12}{-72:-12}=\dfrac{5}{6}\)
Do đó : \(\dfrac{-14}{-21}< \dfrac{-60}{-72}\)
\(c)\dfrac{2135}{13790}và\dfrac{4}{3}\)
Xét phân số : \(\dfrac{2135}{13790}\) ta thấy rằng : \(tử< mẫu\left(2135< 13790\right)\)
\(\Rightarrow\dfrac{2135}{13790}< 1\)
Xét phân số : \(\dfrac{4}{3}có\) : \(tử>mẫu\left(4>3\right)\)
\(\Rightarrow\dfrac{4}{3}>1\)
Do đó : \(\dfrac{2135}{13790}< \dfrac{4}{3}\)
\(d)\dfrac{2022}{2021}và\dfrac{10}{9}\)
Ta thấy rằng : \(\dfrac{2022}{2021}-\dfrac{1}{2021}\text{=}1\)
\(\dfrac{10}{9}-\dfrac{1}{9}\text{=}1\)
Mà : \(\dfrac{1}{9}>\dfrac{1}{2021}\)
\(\Rightarrow\dfrac{2022}{2021}< \dfrac{10}{9}\)
\(e)\dfrac{35}{36}và\dfrac{16}{17}\)
Ta có : \(\dfrac{35}{36}+\dfrac{1}{36}\text{=}1\)
\(\dfrac{16}{17}+\dfrac{1}{17}\text{=}1\)
Mà : \(\dfrac{1}{36}< \dfrac{1}{17}\)
\(\Rightarrow\dfrac{35}{36}>\dfrac{16}{17}\)
\(f)-1,3< -1,2\)
a) Ta có:
\(-\dfrac{11}{12}=\dfrac{1}{12}-1\)
\(-\dfrac{17}{18}=\dfrac{1}{18}-1\)
Mà: \(\dfrac{1}{12}>\dfrac{1}{18}\)
Hay: \(\dfrac{1}{12}-1>\dfrac{1}{18}-1\Rightarrow-\dfrac{11}{12}>-\dfrac{17}{18}\)
b) Ta có:
\(\dfrac{-14}{-21}=\dfrac{2}{3}=\dfrac{4}{6}\)
\(\dfrac{-60}{-72}=\dfrac{5}{6}\)
Mà: \(5>4\Rightarrow\dfrac{-60}{-72}>\dfrac{-14}{-21}\)
c) Ta có:
\(\dfrac{2135}{13790}=\dfrac{61}{394}< 1\) (tử nhỏ hơn mẫu)
\(\dfrac{4}{3}>1\) (tử lớn hơn mẫu)
Ta có: \(\dfrac{61}{394}< \dfrac{4}{3}\Rightarrow\dfrac{2135}{13790}< \dfrac{4}{3}\)
d) Ta có:
\(\dfrac{2022}{2021}=\dfrac{1}{2021}+1\)
\(\dfrac{10}{9}=\dfrac{1}{9}+1\)
Ta thấy: \(\dfrac{1}{2021}< \dfrac{1}{9}\Rightarrow\dfrac{1}{2021}+1< \dfrac{1}{9}+1\)
Hay \(\dfrac{2022}{2021}< \dfrac{10}{9}\)
e) Ta có:
\(\dfrac{35}{36}=1-\dfrac{1}{36}\)
\(\dfrac{16}{17}=1-\dfrac{1}{17}\)
Ta có: \(\dfrac{1}{36}< \dfrac{1}{17}\Rightarrow1-\dfrac{1}{36}>1-\dfrac{1}{17}\)
Hay \(\dfrac{35}{36}>\dfrac{16}{17}\)
f) Ta có: \(1,3>1,2\)
\(\Rightarrow-1,3< -1,2\)
Ta có : 2003/2005 +2/2005=2003/2005+6/6015=1
2128/2134+6/2134=1
Vì 6/6015<6/2134 nên 2003/2005<2128/2134
Vậy 2003/2005<2128/2134
Ta có :2003/2005+2/2005=2003/2005+6/6015=1
2128/2134+6/2034=1
Vì 6/6015<6/2034 nên 2003/2005>2128/2134
2134/2135 > 1439/1440
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