có trong câu hỏi tương tự đấy bạn

mọi người ơi giúp mk với mk đang cần gấp

1)(x^2+3x+1)(x^2+3x+2)-6

Đặt t = x²  + 3x + 1

Khi đó PT có dạng:

t.(t + 1) - 6

= t² + t - 6

= t² - 2t - 3t - 6

= t.(t - 2) + 3.(t - 2)

= (t + 3).(t - 2)

= (x² + 3x + 1 + 3).(x² + 3x + 1 - 2)

= (x² + 3x + 4).(x² + 3x - 1)

\(1\hept{\begin{cases}\left(x^2+3x+2-1\right)\left(x^2+2x+2\right)-6\\\left(t-1\right)\left(t\right)-6\\t^2-t-6\end{cases}}.\) " đặt x^2+3x+2 = t

\(\hept{\begin{cases}t^2-\frac{2t.1}{2}+\frac{1}{4}-\left(\frac{24+1}{4}\right)\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\end{cases}}\)

\(\hept{\begin{cases}\left(t-\frac{1}{2}-\frac{5}{2}\right)\left(t-\frac{1}{2}+\frac{5}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\end{cases}}\)

2)  \(\hept{\begin{cases}\left\{\left(x+1\right)\left(x+7\right)\right\}\left\{\left(x+5\right)\left(x+3\right)\right\}+15\\\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\\t\left(t+8\right)+15\end{cases}}\)  

\(\hept{\begin{cases}t^2+8t+15\\\left(t^2+8t+16\right)-1\\\left(t+4\right)^2-1\end{cases}}\Leftrightarrow\left(t+5\right)\left(t+4\right)\)

\(\hept{\begin{cases}a^3\left(b-c\right)+b^3\left(c-a+b-b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(-c+a-b+b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(a-b\right)-b^3\left(b-c\right)+c^3\left(a-b\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\\\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+ab+c^2\right)\\\left(a-b\right)\left(b-c\right)\left(a^2+2ab+2b^2+c^2\right)\end{cases}}}\)

\(M=5x^2+10y^2-2xy+4x-6y+2\)

        \(=\left(x^2-2xy+y^2\right)+\left(4x^2+4x+1\right)+\left(9y^2-6y+1\right)+1\)

      \(=\left(x-y\right)^2+\left(2x+1\right)^2+\left(3y-1\right)^2+1\ge1\)

vậy \(M\ge N\)

Tại sao có 3 con số 1 vậy pạn??? @Nguyễn_thị_huyền_anh

thêm x;y thuộc z nhé

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

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

vì \(x-y⋮x-y;x,y\in Z\Rightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)⋮x-y\Rightarrow x^8-y^8⋮x-y\)

\(x+y⋮x+y;x,y\in Z\Rightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)⋮x+y\Rightarrow x^8-y^8⋮x+y\)

Cảm ơn bạn nha! Bài bạn làm đúng rồi nhé!