chơi bang bang sv gì solol k

Đặt \(A=x^4-3x^3+6x^2-5x+3\)

Xét trường hợp \(A=\left(x^2+ax+1\right)\left(x^2+bx+3\right)\)

\(A=x^4+bx^3+3x^2+ax^3+abx^2+3ax+x^2+bx+3\)

\(A=x^4+x^3\left(b+a\right)+x^2\left(3+ab+1\right)+x\left(3a+b\right)+3\)

Đồng nhất hệ số ta có:

\(\Rightarrow\hept{\begin{cases}a+b=-3\\3+ab+1=6\\3a+b=-5\end{cases}\Rightarrow\hept{\begin{cases}a+3=-b\\ab=2\\3a+b=-5\end{cases}\Rightarrow}\hept{\begin{cases}a=-1\\b=-2\end{cases}}}\)

Vậy \(x^4-3x^3+6x^2-5x+3=\left(x^2-x+1\right)\left(x^2-2x+3\right)\)

Chúc bn hok tốt ##

\(x^4-3x^3+6x^2-5x+3\)

\(=x^4-2x^3+3x^2-x^3+2x^2-3x+x^2-2x+3\)

\(=\left(x^4-2x^3+3x^2\right)-\left(x^3+2x^2-3x\right)+\left(x^2-2x+3\right)\)

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

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

      \(3xy^2-2xy+12x\)

\(=x\left(3y^2-2y+12\right)\)

Có sai đề không bạn

A= (x+5/2)^2 + 3/4 
nx 
(x+5/2)^2 >=0 suy ra a>= 3/4 
dấu bằng xảy ra khi x+5/2 = 0 <-> x=-5/2 

\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)

\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)\)

\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a\right)\)

\(=\left(a+b+c\right)\left(ab+bc\right)+ca\left(c+a\right)\)

\(=b.\left(a+b+c\right)\left(a+c\right)+ca\left(c+a\right)\)

\(=\left(a+c\right)\left[b.\left(a+b+c\right)+ca\right]\)

\(=\left(a+c\right)\left(ab+b^2+bc+ca\right)\)

\(=\left(a+c\right)\left[a\left(b+c\right)+b\left(b+c\right)\right]\)

\(=\left(a+c\right)\left(b+c\right)\left(a+b\right)\)

\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc\)

\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)+abc\)

\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a+b\right)\)

\(=\left(a+b+c\right)\left(ab+bc+ac\right)\)

Tham khảo nhé~

thank you