c) \(16-x^2+2xy-y^2=\left(4-x+y\right)\left(4+x-y\right)\)

d) \(\left(x-1\right)^2-4\left(2x-3\right)^2=\left(5-3x\right)\left(5x-7\right)\)

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

e) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

c) \(16-x^2+2xy-y^2=\left(4-x+y\right)\left(4+x-y\right)\)

d) \(\left(x-1\right)^2-4\left(2x-3\right)^2=\left(5x-7\right)\left(5-3x\right)\)

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

f) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

=(x-6)(x+1)

`(x+y)^2-4(x+y)`

`=(x+y)(x+y-4)`

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

đặt:\(^{x^2+8x+11=t}\)

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

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

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

tìm có mà link https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-x-1-x-3-x-5-x-7-15-thanh-nhan-tu-faq257547.html

tí mình gửi qua cho 

học tốt

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

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

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

Đặt \(x^2+8x+11=t\)thay vào (1) ta được : 

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

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

\(=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)\)Thay \(t=x^2+8x+11\)vào bt ta được:

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

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

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

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

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

Bài làm

B = ( x + 1 )( x + 3 )( x + 5 )( x + 7 ) + 15 

B = [ ( x + 1 ) ( x + 7 ) ] [ ( x + 3 ) ( x + 5 ) ] + 15

B = [ x² + 7x + x + 7 ] [ x² + 5x + 3x + 15 ] + 15

B = [ x² + 8x + 7 ] [ x² + 8x + 15 ] + 15

Đặt [ x² + 8x + 7 ] [ x² + 8x + 15 ] + 15 = k

=> B = k . ( k + 8 ) + 15

=> B = k² + 8k + 15

=> B = k² + 3k + 5k + 15

=> B = ( k² + 5k ) + ( 3k + 15 )

=> B = [ k( k + 5 ) ] + [ 3( k + 5 ) ]

=> B = ( k + 5 ) ( k + 3 )

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

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

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

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

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

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

# Học tốt #

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)

  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)

A=(x+1)(x+3)(x+5)(x+7)+15=[(x+1)(x+7)][(x+3)(x+5)]+15=(x²+8x+7)(x²+8X+15)+15

Đặt t=x²+8x+7=> A=t²+8t+15=(t+4)²-1=(t+5)(t+3)=(x²+8x+12)(X²+8x+10)=(x+2)(x+6)(x²+8x+10)

vậy...........................................

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)