\(\left(x^2+8x+7\right)\cdot\left(x^2+8x+15\right)+15=\left(x^2+8x+11\right)^2-16+15\)

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

( x² + 8x + 7 )( x² + 8x + 15 ) + 15 (1)

Đặt t = x² + 8x + 7 

(1) <=> t( t + 8 ) + 15

       = t² + 8t + 15

       = t² + 3t + 5t + 15

       = t( t + 3 ) + 5( t + 3 )

       = ( t + 3 )( t + 5 )

       = ( x² + 8x + 7 + 3 )( x² + 8x + 7 + 5 )

       = ( x² + 8x + 10 )( x² + 8x + 12 )

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

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

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

Đặt \(x^2+8x+11=y\)

Từ đó: \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x+11-4\right)\left(x^2+8x+11+4\right)+15\)

\(=\left(y-4\right)\left(y+4\right)+15\)

\(=y^2-16+15\)

\(=y^2-1\)

\(=\left(y-1\right)\left(y+1\right)\)

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

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

x=7

nên x+1=8

\(A=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-5\)

\(=x-5=7-5=2\)

thay x=7

ta có:7^15-8*7^14+887^13-8*7^12+...-8*7^2+8*7-2015

f(7)=7^15-8.7^14+8.7^13-8.7^12+... 
-8.7^2+8.7-5= 
= -7^14+8.7^13-8.7^12+...-8.7^2+8.7-5= 
=7^13-8.7^12+...-8.7^2+8.7-5= 
= -7^12+...-8.7^2+8.7-5= 
=...= -7^2+8.7-5=7-5=2 
Kết quả là 2

sao mà 2 câu giống nhau thế

f(7)=7^15-8.7^14+8.7^13-8.7^12+... 
-8.7^2+8.7-5= 
= -7^14+8.7^13-8.7^12+...-8.7^2+8.7-5= 
=7^13-8.7^12+...-8.7^2+8.7-5= 
= -7^12+...-8.7^2+8.7-5= 
=...= -7^2+8.7-5=7-5=2 
Kết quả là 2

=(x^2+8x)^2+23(x^2+8x)+135

Cái này ko phân tích được nha bạn

\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\\ \Leftrightarrow\left(x^4+8x^3+15x^2+8x^3+64x^2+120x+8x^2+64x+120\right)+15\\ \Leftrightarrow x^4+16x^3+87x^2+184x+135\)

Gọi `A=(x^2+8x+8)(x^2+8x+15)+15`

Đặt `t=x^2+8x+11,5`

`=>A=(t-3,5)(t+3,5)+15=t^2-3,5^2+15=t^2-2,75=(t-sqrt(2,75))(t+sqrt(2,75))=(x^2+8x+11,5-(sqrt11)/2)(x^2+8x+11,5+(sqrt11)/2)=(x^2+8x+(23-\sqrt11)/2)(x^2+8x+(23+\sqrt11)/2)`

vào câu hỏi tương tự

tích nha

Đặt \(t=x^2+8x+11\) và \(A=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\): \(\Rightarrow A=\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)\)

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

x=7 nên x+1=8

B=x^15-x^14(x+1)+x^13(x+1)-...-x^2(x+1)+x(x+1)-5

=x^15-x^15-x^14+x^14+...-x^3-x^2+x^2+x-5

=x-5

=7-5

=2

khó thế

Vì x = 7

\(\Rightarrow\)\(x+1=8\)

\(\Rightarrow\)\(A=x^{15}\)\(-\)\(8x^{14}\)\(+\)\(8x^{13}\)\(-\)\(8x^{12}\)\(+\)... \(-\)\(8x^2\)\(+8x-5\)

\(=\)\(x^{15}\)\(-\left(x+1\right)x^{14}\)\(+\left(x+1\right)x^{13}\)\(-\left(x+1\right)x^{12}\)\(+\)... \(-\)\(\left(x+1\right)x^2\)\(+\left(x+1\right)x-5\)

\(=\)\(x^{15}\)\(-\)\(x^{15}\)\(-\)\(x^{14}\)\(+\)\(x^{14}\)\(+\)\(x^{13}\)\(-\)\(x^{13}\)\(-\)\(x^{12}\)\(+\)... \(-\)\(x^3\)\(-\)\(x^2\)\(+\)\(x^2\)\(+\)\(x\)\(-\)\(5\)

\(\Rightarrow\)\(x=-5\)

\(\Rightarrow\)\(A=7-5=2\)

Vậy  \(A=2\) khi  \(x=7\)