Đề là gì e nhỉ

Gọi 3 số đó là n - 1; n; n + 1 (n > 0)

Theo đề ta có: (n - 1).(n +1) - (n - 1).n  = 9

\(\Rightarrow n^2-1-n^2+n=9\)

\(\Rightarrow n=10\)

Vậy 3 số cần tìm là: 9,10,11

\(12.\left(\dfrac{29292929}{56565656}+\dfrac{66666}{77777}\right)=12.\left(\dfrac{29.1010101}{56.1010101}+\dfrac{6.11111}{7.11111}\right)=12.\left(\dfrac{29}{56}+\dfrac{6}{7}\right)=12.\dfrac{11}{8}=\dfrac{33}{2}\)

101)

\(5\equiv1\left(mod4\right)\)

\(\Rightarrow5^n\equiv1^n=1\left(mod4\right)\)

\(\Rightarrow5^n-1\equiv1-1\left(mod4\right)\)

Hay: \(5^n-1\equiv0\left(mod4\right)\)

Vậy:...

\(\left(x^2+6x+9\right)^3-\left(y^2\right)^3=\left(x^2+6x+9-y^2\right)\left[\left(x^2+6x+9\right)^2+\left(x^2+6x+9\right)y^2+y^4\right]\)

\(=\left[\left(x+3\right)^2-y^2\right]\left\{\left[\left(x^2+6x+9\right)^2+2\left(x^2+6x+9\right)y^2+y^4\right]-\left(x^2+6x+9\right)y^2\right\}\)

\(=\left(x+3-y\right)\left(x+3+y\right)\left[\left(x^2+6x+9+y^2\right)^2-\left(x+3\right)^2y^2\right]\)

\(=\left(x+3-y\right)\left(x+3+y\right)\left[\left(x^2+6x+9+y^2\right)-\left(x+3\right)y\right]\left(x^2+6x+9+y^2\right)+\left(x+3\right)y\)

\(=\left(x+3-y\right)\left(x+3+y\right)\left(x^2+6x+9+y^2-xy-3y\right)\left(x^2+6x+9+y^2+xy+3y\right)\)

mình ms 2k7 thôi nhaa đừng xưng hô vậy chứ :((