\(3^{135}+2^{135}+3^{133}+2^{134}\)

\(=\left(3^{135}+3^{133}\right)+\left(2^{135}+2^{134}\right)\)

\(=3^{133}\cdot\left(3^2+1\right)+2^{133}\cdot\left(4+1\right)\)

\(=3^{133}\cdot10+2^{133}\cdot5\)

\(=5\cdot2\cdot\left(3^{133}+2^{132}\right)\)

\(=10\cdot\left(3^{133}+2^{132}\right)\)

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=2136x(2134+2)

a=2136x2134+2136x2

b=2134x(2136+2)

b=2134x2136x2134x2

vì 2136x2>2134x2

nên 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: \(1,3>1,2\)

\(\Rightarrow-1,3< -1,2\)

2003/2005 lớn hơn

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

bài làm trên sai mà bạn