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Ta có x(x+3)(x+2)(x+5)+9= x(x+5).(x+2)(x+3) +9= (x2+5x)(x2+5x+6)+9
Đặt x2+5x+3=a ta được
(a-3).(a+3)+9= a2-9+9=a2
Thay x2+5x+3 vào biểu thức trên ta được
(x2+5x+3)2
Vậy x(x+3)(x+2)(x+5)= (x2+5x+3)2
\(x\left(x+3\right)\left(x+2\right)\left(x+5\right)+9\)
\(=\left(x^2+5x\right)\left(x^2+5x+6\right)+9\)
\(=\left[\left(x^2+5x+3\right)-3\right]\left[\left(x^2+5x+3\right)+3\right]+9\)
\(=\left(x^2+5x+3\right)^2-9+9\)
\(=\left(x^2+5x+3\right)\)
M = x9 - x7 + x6 - x5 - x4 + x3 - x2 + 1
= ( x9 - x7 ) + ( x6 - x4 ) - ( x5 - x3 ) - ( x2 - 1 )
= x7( x2 - 1 ) + x4( x2 - 1 ) - x3( x2 - 1 ) - ( x2 - 1 )
= ( x2 - 1 )( x7 + x4 - x3 - 1 )
= ( x - 1 )( x + 1 )[ x4( x3 + 1 ) - ( x3 + 1 ) ]
= ( x - 1 )( x + 1 )( x3 + 1 )( x4 - 1 )
= ( x - 1 )( x + 1 )( x + 1 )( x2 - x + 1 )( x2 - 1 )( x2 + 1 )
= ( x + 1 )2( x - 1 )( x2 - x + 1 )( x - 1 )( x + 1 )( x2 + 1 )
= ( x + 1 )3( x - 1 )2( x2 + 1 )( x2 - x + 1 )
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)(1)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)-15=t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2=\left(t+5\right)\left(t-3\right)\)
\(=\left(x^2+5x+9\right)\left(x^2+5x+1\right)\)
b) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)
\(=\left(3x-4\right)\left(x+14\right)\)
Ta có:
\(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)
\(=\left(x^9-x^8\right)+\left(x^8-x^7\right)-\left(x^6-x^5\right)-\left(2x^5-2x^4\right)-\left(x^4-x^3\right)+\left(x^2-x\right)+\left(x-1\right) \)
\(=x^8.\left(x-1\right)+x^7.\left(x-1\right)-x^5.\left(x-1\right)-2x^4.\left(x-1\right)-x^3\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^8+x^7-x^5-2x^4-x^3+x+1\right)\)
x^9 + x^8 + x^7 - x^3 + 1
= x^7 ( x^2 + x + 1 ) - ( x^3 - 1 )
= x^7 ( x^2 + x + 1 ) - ( x - 1 )(x^2 + x + 1 )
= ( x^7 - x + 1 )(x^2 + x + 1 )
Ta có \(f\left(x\right)=\left(x+1\right)\left(x-3\right)\left(x+5\right)\left(x+9\right)+256\)
\(f\left(x\right)=\left(x+1\right)\left(x+5\right)\left(x-3\right)\left(x+9\right)+256\)
\(f\left(x\right)=\left(x^2+6x+5\right)\left(x^2+6x-27\right)+256\)
\(f\left(x\right)=\left[\left(x^2+6x-11\right)^2-256\right]+256\)
\(f\left(x\right)=\left(x^2+6x-11\right)^2\)