n^9 + 1 = 1n^9

\(n^7+n^2+1\)

\(=\left(n^7-n^6+n^4-n^3+n^2\right)\)\(+\left(n^6-n^5+n^3-n^2+n\right)\)

\(+\left(n^5-n^4+n^2-n+1\right)\)

\(=n^2\left(n^5-n^4+n^2-n+1\right)\)\(+n\left(n^5-n^4+n^2-n+1\right)\)

\(+\left(n^5-n^4+n^2-n+1\right)\)

\(=\left(n^2+n+1\right)\left(n^5-n^4+n^2-n+1\right)\)

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

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

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

n³ -n² -7n +1=n²(n-1) - 7(n-1)-6=0 <=>(n-1).(n²-7)=6

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

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

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

Đặt: \(a^2+8a+11=t\), khi đó pt trở thành:

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

\(=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\\ =\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

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

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

Đặt \(t=a^2+8a+7\) khi đó A thành:

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

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

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

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

Ta có:

\(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

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

\(\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

Đặt \(a^2+8a+7=t\)

\(\Rightarrow t\left(t+8\right)+15\)

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

\(=t\left(t+3\right)+5\left(t+3\right)\)

\(=\left(t+3\right)\left(t+5\right)\)

\(\Rightarrow\left[\left(a^2+8a+7\right)+3\right]\left[\left(a^2+8a+7\right)+5\right]\)

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

\(=\left(a^2+8a+10\right)\left(a^2+2a+6a+12\right)\)

\(=\left(a^2+8a+10\right)\left[a\left(a+2\right)+6\left(a+2\right)\right]\)

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

Đặt \(M=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

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

\(M=\left(a^2+7a+a+7\right)\left(a^2+5a+3a+15\right)+15\)

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

Đặt \(p=a^2+8a+11\)

\(\Rightarrow M=\left(p-4\right)\left(p+4\right)+15\)

\(\Rightarrow M=p^2-16+15\)

\(\Rightarrow M=p^2-1\)

\(\Rightarrow M=\left(p-1\right)\left(p+1\right)\)

Thay \(p=a^2+8a+11\)vào M, ta có :

\(M=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)\)

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

(a+1)(a+7)(a+3)(a+5)+15
=(a²+8a+7)(a²+8a+15)+15
=(a²+8a+11-4)(a²+8a+11+4)+15
=(a²+8a+11)²-4²+15
=(a²+8a+11)²-1
=(a²+8a+11-1)(a²+8a+11+1)
=(a²+8a+10)(a²+8a+12)
 

Rút gọn biểu thức sau:A=(2x-3)(2x+3)-(x+5)²-(x-1)(x+2)

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

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

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

Đặt : \(a^2+8+11=t\) khi đó pt trở thành :

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

\(=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)\)

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

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

Chúc bạn học tốt !!!

A = (a+1)(a+3)(a+5)(a+7) + 15

A = [ (a+1) (a+7)] [(a+3) (a+5)] + 15

A= ( a² + 8a + 7)( a² + 8a + 15 ) + 15                 (*)

         Đặt a² + 8a + 7 = t

=> A = t.(t+8) + 15

     A = t² + 8t + 15

     A = t² + 3t + 5t + 15

     A = ( t +3).(t+5)

  Thay   A = ( t +3).(t+5) vào (*)

=> A = ( a² + 8a + 7 + 3).( a² + 8a + 7 + 5)

    A = ( a² + 8a + 10).( a² + 8a + 12 )

     A = ( a² + 8a + 10).( a² + 6a + 2a + 12 )

       A = ( a² + 8a + 10) ( a+6)(a+2)

a=(a+1)(a+3)(a+5)(a+7)+15

a=[ (a+1)(a+7) ] [(a+3)(a+5)] +15

a=(a²+8a+7)(a²+8a+15)+15

        Đặt a²+8a+7=t

         a=t.(t+8)+15

         a=t²+8t+15

          a=t²+3t+5t+15

         a=(t+3).(t+5)

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