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(x - 5)2 - 4(x - 3)2 + 2(2x - 1)(x - 5) + (2x - 1)2
= [(x - 5)2 + 2(2x - 1)(x - 5) + (2x - 1)2) - [2(x - 3)]2
= (x - 5 + 2x - 1)2 - (2x - 6)2
= (3x - 6)2 - (2x - 6)2
= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)
( x - 5 )2 - 4( x - 3 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2
= [ ( x - 5 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2 ] - 22( x - 3 )2
= ( x - 5 + 2x - 1 )2 - ( 2x - 6 )2
= ( 3x - 6 )2 - ( 2x - 6 )2
= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )
= x( 5x - 12 )
\(=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+3=\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
\(=\left(x+1\right)^2\left(x^2+2x+3\right)\)
\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)
\(=x^8+2x^5-2x^4+x^2-2x-100+10x^5+25x^2-10x+1\)
\(=x^8+12x^5-2x^4+36x^2-12x-99\)
\(=x^8+6x^5+9x^4+6x^5+36x^2+54x-11x^4-66x-99\)
\(=x^4\left(x^4+6x+9\right)+6x\left(x^4+6x+9\right)-11\left(x^4+6x+9\right)\)
\(=\left(x^4+6x+9\right)\left(x^4+6x-11\right)\)
\(2\left(x^2+x+1\right)^2-\left(2x+1\right)^2-\left(x^2+2x\right)^2\)
\(=2.\left[x^4+x^2+1+2x^3+2x+2x^2\right]-\left(4x^2+4x+1\right)-\left(x^4+4x^3+4x^2\right)\)
\(=x^4-2x^2+1=\left(x^2-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)
Chúc bạn học tốt.
Đặt \(x^2-2x+4=a\)
Khi đó \(\left(x^2-2x+3\right)\left(x^2-2x+5\right)-8=\left(a-1\right)\left(a+1\right)-8\)
\(=a^2-1-8\)
\(=a^2-9\)
\(=\left(a-3\right)\left(a+3\right)\)
\(=\left(x^2-2x+4-3\right)\left(x^2-2x+4+3\right)\)
\(=\left(x^2-2x+1\right)\left(x^2-2x+7\right)\)
\(=\left(x-1\right)^2\left(x^2-2x+7\right)\)
Khó quá , bó tay
\(\left(x^2-2x+2\right)^4-20x^2\left(x^2-2x+2\right)+64x^4\)
\(=\left[\left(x^2-2x+2\right)^2\right]^2-2.\left(x^2-2x+2\right)^2.10x^2+\left(10x^2\right)^2-36x^4\)
\(=\left[\left(x^2-2x+2\right)^2-10x^2\right]^2-\left(6x^2\right)^2\)\(=\left[\left(x^2-2x+2\right)^2-4x^2\right]\left[\left(x^2-2x+2\right)^2-16x^2\right]\)
\(=\left(x^2-2x+2+2x\right)\left(x^2-2x+2-2x\right)\left(x^2-2x+2-4x\right)\left(x^2-2x+2+4x\right)\)
\(=\left(x^2+2\right)\left(x^2-4x+2\right)\left(x^2-6x+2\right)\left(x^2+2x+2\right)\)