áoima 

Giúp tôi vs nhé mn tôi phải nộp gấp trong ngày mai 😕😕😕

\(\left(x+1\right)^3=125\)

\(\left(x+1\right)^3=\sqrt[3]{125}\)

\(\left(x+1\right)=5\)

\(x=5-1\)

\(x=4\)

\(2^x+2^{x+3}=576\)

\(2^x\left(1+2^3\right)=576\)

\(2^x=576:9\)

\(2^x=64\)

\(2^x=2^6\)

\(\Rightarrow x=6\)

\(2^x+2^{x+3}=576\)

\(\Rightarrow2^x+2^x.2^3=24^2\)

\(\Rightarrow2^x\left(1+2^3\right)=24^2\)

\(\Rightarrow2^x+3^2=24^2\)

\(\Rightarrow2^x=\left(24:3\right)^2\)

\(\Rightarrow2^x=8^2\)

\(\Rightarrow2^x=\left(2^3\right)^2\)

\(\Rightarrow2^x=2^6\)

\(\Rightarrow x=6\)

a, \(\left(x+1\right)^3=125\Rightarrow\left(x+1\right)^3=5^3\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=5-1=4\)

b, \(2^x+2^{x+3}=576\)

\(\Rightarrow2^x+2^x.2^3=576\)  \(\Rightarrow2^x\left(1+2^3\right)=576\) \(\Rightarrow2^x.9=576\)

\(\Rightarrow2^x=\frac{576}{9}=64\)  \(\Rightarrow2^x=2^6\) \(\Rightarrow x=6\)

đặc trưng là cách nhau 1 đơn vị

có 20 phần tử chẵn

có 15 phần tử lẻ

Gọi \(n=\left(a,c\right)\) \(\Rightarrow\left\{{}\begin{matrix}a=na_1\\c=nc_1\end{matrix}\right.\)

+ \(ab=cd\Rightarrow na_1b=nc_1d\)

\(\Rightarrow a_1b=c_1d\) (1)

\(\Rightarrow b⋮c_1\Rightarrow b=mc_1\)

Thay \(b=mc_1\) vào (1) ta có :

\(a_1mc_1=c_1d\Rightarrow d=ma_1\)

Do đó : \(a^{2018}+b^{2018}+c^{2018}+d^{2018}\)

\(=\left(na_1\right)^{2018}+\left(mc_1\right)^{2018}+\left(nc_1\right)^{2018}+\left(ma_1\right)^{2018}\)

\(=a_1^{2018}\left(m^{2018}+n^{2018}\right)+c_1^{2018}\left(m^{2018}+n^{2018}\right)\)

\(=\left(a_1^{2018}+c_1^{2018}\right)\left(m^{2018}+n^{2018}\right)\)

=> đpcm