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1^3-3^5-(-3^5)+1^64-2^9-1^36+1^15
=1+(-3^5+3^5)+1-2^9-1+1
=2-2^9
=-510
1: so sánh 2016/2017+2017/2018
vì 2016/2017 > 1/2017 >1/2018 =
> 2016/2017+2017/2018 >1/2018+2017/2018=1
vậy .....
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
\(a,\left|2x-5\right|=1\)
\(\Rightarrow\orbr{\begin{cases}2x-5=1\\2x-5=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\2x=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
b, đề thiếu
Câu 1 :
\(\dfrac{-25}{37}\&\dfrac{-20}{31}\)
Ta thấy \(\dfrac{-25}{37}< \dfrac{-20}{37}\)
mà \(\dfrac{-20}{37}< \dfrac{-20}{31}\)
\(\Rightarrow\dfrac{-25}{37}< \dfrac{-20}{31}\)
Câu 2 :
\(\dfrac{2}{3}\&\dfrac{5}{7}\)
\(\dfrac{2}{3}:\dfrac{5}{7}=\dfrac{2}{3}.\dfrac{7}{5}=\dfrac{14}{15}< 1\)
\(\Rightarrow\dfrac{5}{7}>\dfrac{2}{3}\) Câu 3 : \(\dfrac{8}{13}\&\dfrac{5}{7}\)Ta thấy \(\dfrac{8}{13}:\dfrac{5}{7}=\dfrac{8}{13}.\dfrac{7}{5}=\dfrac{56}{65}< 1\)
\(\Rightarrow\dfrac{8}{13}< \dfrac{5}{7}\)3: \(=\dfrac{13}{5}\left(-\dfrac{3}{14}+\dfrac{2}{5}+\dfrac{-11}{14}+\dfrac{3}{5}\right)\)
=0
đề dài vậy
bạn làm đc ko??