\(a^7+a^2+1=a^7-a+a^2+a+1=a\left(a^3-1\right)\left(a^3+1\right)+\left(a^2+a+1\right)\)

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

\(=\left(a^2+a+1\right)\left[a\left(a-1\right)\left(a^3+1\right)+1\right]=\left(a^2+a+1\right)\left(a^5-a^4+a^2-a+1\right)\)

a^5+a+1=a^5-a^2+(a^2+a+1)

=a^2(a^3-1)+(a^2+a+1)

a^2(a-1)(a^2+a+1)+(a^2+a+1)

(a^2+a+1)(a^3-a^2+1)

(a^2+a+1)(

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

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

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

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 \(a^2+8a+7=t\)

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

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

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

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

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

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

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

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

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

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

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

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

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)

Hok tok

=a^4(a+1)+a^2(a+1)+(a+1)

=(a+1)(a^4+a^2+1)

a⁵+a⁴+a³+a²+a+1

=a⁴(a+1)+a²(a+1)+(a+1)

=(a+1)(a⁴+a²+1)

 1)a)x(x+4)(x-4)-(x2+1)(x2-1) 
=x(x2-16)-(x^4-1)=-x^4+x^3-16x+1 
b)4(x-1)(x+1)-(x+2)(x+5)-3(x-1)(x+2) 
=4(x2-1)-(x2+7x+10)-3(x2+x-2) 
=4x2-4-x2-7x-10-3x2-3x+6 
=-10x-8 
2)a)x2-2x-4y2-4y 
=(x2-4y2)-(2x+4y) 
=(x-2y)(x+2y)-2(x+2y) 
=(x+2y)(x-2y-2) 
b)x2-x-2=x2-2x+x-2 
=x(x-2)+(x-2)=(x-2)(x+1) 
c)a5+a+1=a5+a4+a3-a4-a3-a2+a2+a+1 
=a3(a2+a+1)-a2(a2+a+1)+(a2+a+1) 
=(a2+a+1)(a3-a2+1) 
3)a)x3-4x2-8x+8 
=(x3+8)-(4x2+8x) 
=(x+2)(x2-2x+4)-4x(x+2) 
=(x+2)(x2-6x+4) 
b)4x4+1 
=4x4+4x2+1-4x2 
=(2x2+1)^2 -(2x)^2 
=(2x2+1+2x)(2x2+1-2x) 
c)x2(x2+4)-x2+4 
=x4-x2+4x2+4 
=x4+4x2+4-x2 
=(x2+2)^2-x^2 
=(x2-x+2)(x2+x-2) 
=(x2-x+2)(x2+2x-x-2) 
=(x2-x+2)(x+2)(x-1) 
4)a)Bạn đặt phép chia (2x2+ax+1)cho(x-3)thì ta được số dư là :3a+19(bài dài quá nên mình không muốn đánh bạn tự chia nhé).mà để phép chia dư 4 thì 3a+19=4 =>a=-5 
vậy a=-5 
b)ta đặt:g(x)= ax5+5x4-x=(x-1).f(x) 
g(1)=a+4=0.f(x)=0 =>a=-4 
Bạn làm như anh Hồ Công Tử là tốt lắm rồi 
5)a)(5-2x)(2x+7)=4x2-25 
<=>(5-2x)(2x+7)+(25-4x2)=0 
<=>(5-2x)(2x+7)+(5-2x)(5+2x)=0 
<=>(5-2x)(2x+7+5+2x)=0 
<=>(5-2x)(4x+12)=0 
<=>5-2x=0 hoặc 4x+12=0 
<=>x=2,5 hoặc x=-3 
b)x3+27+(x+3)(x-9)=0(bạn hãy chú ý đến 3^3=27và a^3+b^3=(a+b)(a2-ab+b2)) 
<=>(x+3)(x2-3x+9)+(x+3)(x-9)=0 
<=>(x+3)(x2-2x)=0 
<=>x(x+3)(x-2)=0 
<=>x=0 hoặc x+3=0 hoặc x-2=0 
<=>x=0 hoặc x=-3 hoặc x=2 
c)x2+3x-18=0 
<=>x2+6x-3x-18=0 
<=>x(x+6)-3(x+6)=0 
<=>(x+6)(x-3)=0 
<=>x+6=0 hoặc x-3=0 
<=>x=-6 hoặc x=3 

k mình nha mấy bạn

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

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

=(a²+8a+7)(a²+8a+15)+15

Đặt y=a²+8a+7 ta được :

y(y+8)+15=y² + 8y +15

=y² +3y+5y+15

=y(y+3) +5(y+3)

=(y+3)(y+5)

thay y=a²+8a+7 ta được 

(a²+8a+7+3)(a²+8a+7+5)

=(a²+8a+10)(a²-2a-6a+12)

=(a²+8a+10)[a(a-2)-6(a-2)]

=(a²+8a+10)(a-2)(a-6)

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)

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