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mik làm phần b nhé vì phần a có người làm rồi
x4-5x2+4=x4-x2-4x2+4=(x4-x2)-(4x2-4)
=x2(x2-1)-4(x2-1)
=(x2-1)(x2-4)
=(x-1)(x+1)(x-2)(x+2)
\(x^4-5x^2+4\)
=\(\left(x^4-x^2\right)-\left(4x^2-4\right)\)
=\(x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
=\(\left(x^2-4\right)\left(x^2-1\right)\)
\(x^2+5x-6\)
=\(x^2-x+6x-6\)
=\(x\left(x-1\right)+6\left(x-1\right)\)
=\(\left(x+6\right)\left(x-1\right)\)
2 câu cuối làm tương tự nha câu 2 nha
\(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
a) \(x^4-9x^2\)
\(=x^2\left(x^2-9\right)\)
\(=x^2\left(x-3\right)\left(^{ }x+3\right)\)
b) \(3x^2-12x+12\)
\(=3x\left(x^2-4x+4\right)\)
\(=3x\left(x-2\right)^2\)
c) \(x^2+5x+6\)
\(=x^2+3x+2x+6\)
\(=x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x+2\right)\)
x4 - 9x2
= x4 - ( 3x )2
= ( x2 - 3x ) ( x2 + 3x )
b) 3x3 - 12x2 + 12x
= 3x3 - 6x2 - 6x2 + 12x
= 3x2( x - 2 ) - 6x ( x - 2 )
= ( 3x2 - 6x ) ( x - 2 )
= 3x ( x - 2 ) ( x - 2 )
= 3x ( x- 2 )2
c) x2 + 5x + 6
= x2 + 2x + 3x + 6
= x ( x + 2 ) + 3 ( x + 2 )
= ( x + 3 ) ( x + 2 )
a, \(\left(4x+5\right)^2=\left(4x+5\right)\left(4x+5\right)=\left[\left(4x+5\right)4x\right]+\left[\left(4x+5\right)5\right]=4x^2+20x+25\)
b, \(\left(5x-2\right)^2=\left(5x-2\right)\left(5x-2\right)=\left[\left(5x-2\right)5x-\left(5x-2\right)2\right]=5x^2-10x+25\)
b, \(8^2-12x^2=\left(8^2-12x^2\right)\left(8^2+12x^2\right)\)
đúng ko :)
@No name: Bị sai rồi nhé, a,b,c sai hết :>
a) ( 4x + 5 )2
= ( 4x )2 + 2.4x.5 + 52
= 16x2 + 40x + 25
b) ( 5x - 2 )2
= ( 5x )2 - 2.5x.2 + 22
= 25x2 - 20x + 4
c) 82 - 12x2
= 64 - 12x2
= ( V8 - V12x )( V8 + V12x )
4x3+4x4−x2−x
=4x3(x+1)−x(x+1)
=(x+1)(4x3−1)
ĐÂY NHÉ. T.I.C.K MÌNH VỚI
\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
\(x^4-5x^2+4=x^4-x^2-4x^2+4=x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
Ta có : x4 - 5x2 + 4
= x4 - x2 - 4x2 + 4
= x2(x2 - 1) + (4x2 - 4)
= x2(x2 - 1) + 4(x2 - 1)
= (x2 - 1)(x2 + 4)