ko đăng linh tinh nhé bạn

mk chơi nè

có j thì bạn lập sever để mk vào sinh tồn cùng nhá

(A+B+C+D)²   

^{phân tích ra hằng dẳng thức}

^{(A+B+C+D)^2=A^2+B^2+C^2+D^2+2AB+2AC+2AD+2BC+2BD+2CD}

Chuc hoc tot!

mk nx nha

\(\left(a+b+c+d\right)^2\)

\(=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\)

Tham khảo nhé~

de minh lop 12A1

(A+B+C+D)²   

^{phân tích ra hằng dẳng thức}

(A+B+C+D)^2=A^2+B^2+C^2+D^2+2AB+2AC+2AD+2BC+2BD+2CD

Khong dang cau hoi linh tinh nha!

có tui

\(x^8+x^4+1=\left(x^8+2x^4+1\right)-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)

Câu c tương tự

Câu b xin thêm thời gian 

tham khảo nhé~

\(x^{10}+x^5+1=\left(x^{10}-x\right)+\left(x^5-x^2\right)+\left(x^2+x+1\right)\)

\(=x\left(x^9-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3-1\right)\left(x^6+x^3+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)\left(x^6+x^3+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^6+x^3+1\right)+x^2\left(x-1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left[\left(x^2-x\right)\left(x^6+x^3+1\right)+x^3-x^2+1\right]\)

\(=\left(x^2+x+1\right)\left(x^8+x^5+x^2-x^7-x^4-x+x^3-x^2+1\right)\)

\(=\left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)

vào đây tham  khảo nè https://vn.answers.yahoo.com/question/index?qid=20090610061522AAlSvoW

ko bt vô

          \(x^{12}+1=\left(x^4\right)^3+1=\left(x^4+1\right)\left(x^8-x^4+1\right)\)