\(bc\left(b+c\right)+ca\left(c-a\right)-ab\left(a+b\right)\\ =bc\left(b+c\right)+ca\left(c-a\right)-ab\left(a+b\right)-\left(abc-abc\right)\\ =b^2c+bc^2+c^2a-ca^2-a^2b-ab^2-abc+abc\\ =\left(b^2c+bc^2+c^2a+abc\right)-\left(ca^2+a^2b+ab^2+abc\right)\\ =c\left(b^2+bc+ca+ab\right)-a\left(ca+ab+b^2+bc\right)\\ =\left(c-a\right)\left(ca+ab+b^2+bc\right)\\ =\left(c-a\right)\left[\left(ca+ab\right)+\left(b^2+bc\right)\right]\\ =\left(c-a\right)\left[a\left(b+c\right)+b\left(b+c\right)\right]\\ =\left(c-a\right)\left(a+b\right)\left(b+c\right)\)

Sau này sẽ cố gắng. -.-

Theo bài ra ta có : \(x=79\Rightarrow\left\{{}\begin{matrix}1969=x+1890\\80=1+x\end{matrix}\right.\left(1\right)\)

\(\text{Thay }\left(1\right)\text{ biểu thức }E=1969-80x+80x^2-80x^3+80x^4-...+80x^{1968}-x^{1969}\)

\(E=1890+x-\left(1+x\right)x+\left(1+x\right)x^2-\left(1+x\right)x^3+...+\left(1+x\right)x^{1968}-x^{1969}\)

\(E=1890+x-x-x^2+x^2+x^3-x^3-x^4+...+x^{1968}+x^{1969}-x^{1969}\)

\(E=1890\)

Vậy giá trị của biểu thức \(E=1969-80x+80x^2-80x^3+80x^4-...+80x^{1968}-x^{1969}\) tại \(x=79\) là \(1890\)

Đây bạn nhé

https://www.google.com.vn/url?sa=t&rct=j&q=&esrc=s&source=video&cd=3&cad=rja&uact=8&ved=0ahUKEwjd_Nfui-3VAhUMMY8KHeU5Dw8QtwIIKjAC&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3D2d4cOD-n1QA&usg=AFQjCNHL1wFqDknJ4lSUMaSxhJxBQXZnyQ

Điểm danh

Viết phân số thích hợp vào chỗ chấm:

1 c m 2   =   … … d m 2 8 c m 2   =   … … d m 2 27 c m 2   =   … … d m 2

Tìm GTNN nhé

\(A=x^2+2y^2+2xy+2y\\ A=\left(x^2+2xy+y^2\right)+y^2+2y+1-1\\ A=\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)-1\\ A=\left(x+y\right)^2+\left(y+1\right)^2-1\)

\(\text{ Ta có : }\left(x+y\right)^2\ge0\\ \left(y+1\right)^2\ge0\\ \Rightarrow\left(x+y\right)^2+\left(y+1\right)^2\ge0\\ A=\left(x+y\right)^2+\left(y+1\right)^2-1\ge-1\)

\(\text{Dấu }"="\text{ xảy ra khi : }\left\{{}\begin{matrix}\left(y+1\right)^2=0\\\left(x+y\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)

Vậy \(A_{\left(Min\right)}=-1\) khi \(x=1\) và \(y=-1\)