\(=\left(x^2+8x+15\right)\left(x^2+8x+7\right)+15\)

đặt:\(^{x^2+8x+11=t}\)

ta co \(\left(t+4\right)\left(t-4\right)+15=t^2-16+15=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)\Rightarrow\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)\)

\(\Rightarrow\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(C=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) \(\left(1\right)\)

Đặt \(x^2+8x+11=t\) , khi đó

\(\left(1\right)\Leftrightarrow\left(t-4\right)\left(t+4\right)+15\)

\(=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\\ =\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

\(C=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(t=x^2+8x+7\) thì C trở thành:

\(t\left(t+8\right)+15=t^2+8t+15\)

\(t^2+3t+5t+15=t\left(t+3\right)+5\left(t+3\right)\)

\(=\left(t+5\right)\left(t+3\right)=\left(x^2+8x+7+5\right)\left(x^2+8x+7+3\right)\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

\(C=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(C=\left(x^2+8+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(x^2+8x+7=k\)

\(\Rightarrow C=k\left(k+8\right)+15=k^2+8k+15\)

\(\Rightarrow C=k^2+3k+5k+15\)

\(\Rightarrow C=k\left(k+3\right)+5\left(k+3\right)\)

\(\Rightarrow C=\left(k+3\right)\left(k+5\right)\)

\(\Rightarrow C=\left(x^2+8x+7+6\right)\left(x^2+8x+7+3\right)\)

\(\Rightarrow C=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(\Rightarrow C=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

a)x^5+x+1

=x⁵-x²+x²+x+1

=x²(x³-1)+x²+x+1

=x²(x+1)(x²+x+1)+x²+x+1

=(x²+x+1)(x³+x²+1)

b)(x+1)(x+3)(x+5)(x+7)+15

=(x²+8x+7)(x²+8x+15)+15

Đặt x²+8x+7=t

=> t(t+8)+15=t²+8t+15

=(t+3)(t+5)

=(x²+8x+10)(x²+8x+12)

Bài 1 :

\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

Bài 2 :

 \(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)

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

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

Tick đúng nha 

X^2-6+8

Đặt \(A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

Ta có : \(A=\left[\left(x+1\right)\left(x+7\right)\right].\left[\left(x+3\right)\left(x+5\right)\right]+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(t=x^2+8x+11\) , suy ra \(A=\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)\)

\(\Rightarrow A=\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

f(x) = (x+1)(x+3)(x+5)(x+7)+15

        = (x+1)(x+7)(x+3)(x+5)+15

        = (x²+7x+x+7)(x²+5x+3x+15)+15

        = (x²+8x+7)(x²+8x+15)+15

        Đặt X=x²+8x+11

   f(x) = (X-4)(X+4)+15

         = X²-16+15

         = X²-1²

         = (X-1)(X+1)

=> f(x)= (x²+8x+11-1)(x²+8x+11+1)

     f(x) = (x²+8x+10)(x²+8x+12)

Đến đây là vẫn còn phân tích được nhưng không dùng phương pháp đặt biến phụ:

     f(x) = (x²+8x+10)(x²+8x+12)

           = (x²+8x+10)[(x²+2x)+(6x+12)]

           = (x²+8x+10)[x(x+2)+6(x+2)]

           = (x+2)(x+6)(x²+8x+10)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)