\(1,\) Ta có \(2022\equiv1\left(mod47\right)\)
\(\Rightarrow2022^{2021}\equiv1\left(mod47\right)\)
Vậy \(2022^{2021}:47\) dư 1
\(2,\) Thay \(x=1\) vào nhị thức, ta được \(\left(5x-6\right)^{2021}=\left(-1\right)^{2021}=-1\)
Vậy tổng các hệ số là \(-1\)
\(1,\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)
Đặt \(a+b-2c=x;b+c-2a=y;c+a-2b=z\Leftrightarrow z=x+y\), pt trở thành:
\(x^3+y^3+z^3\\ =\left(x+y\right)^3-3xy\left(x+y\right)+z^3\\ =-z^3-3xy\left(-z\right)+z^3\\ =3xyz\\ =3\left(a+b-2c\right)\left(b+c-2a\right)\left(a+c-2b\right)\)
\(2,\left(a+b+c\right)^3+\left(a-b-c\right)^3+\left(b-c-a\right)^3+\left(c-a-b\right)^3\\ =8a^3-3\left(a+b+c\right)\left(a-b-c\right)\cdot2a-8a^3-3\left(b-c-a\right)\left(c-a-b\right)\left(-2a\right)\\ =-6a\left\{a^2-\left(b+c\right)^2-\left[\left(-a\right)^2-\left(b-c\right)^2\right]\right\}\\ =-6a\left[a^2-a^2+\left(b-c\right)^2-\left(b+c\right)^2\right]\\ =-6a\left(b-c+b+c\right)\left[b-c-\left(b+c\right)\right]=24abc\)
