Phân tích đa thức thành nhân tử: \(a^3+4a^2-19a+24\)
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E = x^3 - 3x^2 + 7x^2 - 21x - 8x + 24
= x^2 ( x- 3 ) + 7x ( x- 3 ) - 8 ( x- 3 )
= ( x- 3 )(x^2 + 7x - 8 )
= ( x- 3 )[ x^2 + 8x - x - 8 )
= ( x -3 ) [ x(x + 8 ) - ( x + 8 ) ]
= ( x- 3 )( x - 1 )( x + 8)
=a3-3a2+7a2-21a-8a+24
=a2(a-3)+7a(a-3)-8(a-3)
=(a-3)(a2+7a-8)
=(a-3)(a2-a+8a-8)
=(a-3)(a+8)(a-1)
\(a^3+4a^2+4a+3\)
\(=a^3+a^2+3a^2+3a+a+3\)
\(=\left(a^3+a^2+a\right)+\left(3a^2+3a+3\right)\)
\(=a\left(a^2+a+1\right)+3\left(a^2+a+1\right)\)
\(=\left(a+3\right)\left(a^2+a+1\right)\)
\(a^3+4a^2-7a-10\)
\(=\left(a^3+5a^2\right)-\left(a^2+5a\right)-\left(2a+10\right)\)
\(=a^2\left(a+5\right)-a\left(a+5\right)-2\left(a+5\right)\)
\(=\left(a^2-a-2\right)\left(a+5\right)\)
\(=\left(a^2-2a+a-2\right)\left(a+5\right)\)
\(=\left[a\left(a-2\right)+\left(a-2\right)\right]\left(a+5\right)\)
\(=\left(a+1\right)\left(a-2\right)\left(a+5\right)\)
\(a.25^2-4a^2+12ab-9b^2\\ =25^2-\left(4a^2+12ab-9b^2\right)\\ =25^2-\left(2a-3b\right)^2\\ =\left(25-2a+3b\right)\left(25+2a-3b\right)\\ b.x^3+x^2y-xy^2-y^3\\ =x^2\left(x+y\right)-y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-y^2\right)\\ =\left(x+y\right)\left(x+y\right)\left(x-y\right)\\ =\left(x+y\right)^2\left(x-y\right)\)
a: Ta có: \(25x^2-4a^2+12ab-9b^2\)
\(=25x^2-\left(2a-3b\right)^2\)
\(=\left(5x-2a+3b\right)\left(5x+2a-3b\right)\)
b: Ta có: \(x^3+x^2y-xy^2-y^3\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
\(4a^2-4a+1-4b^2\)
<=>\(\left(2a-1\right)^2-4b^2\)
<=>\(\left(2a-1+2b\right)\left(2a-1-2b\right)\)
\(4a^2-4a+1-4b^2\)
\(=\left(2a-1\right)^2-4b^2\)
\(=\left(2a-1+2b\right)\left(2a-1-2b\right)\)
4a2b2 + 36a2b3 + 6ab4
= 2ab2(2a + 18ab + 3b2)
4a2b3 - 6a3b2
= 2a2b2(2b - 3a)
a2 - 4a + 3
= a2 - a - 3a + 3
= a.(a - 1) - 3.(a - 1)
= (a - 1).(a - 3)
= \(a^2-3a-a+3\)= \(a\left(a-3\right)-\left(a-3\right)\)= \(\left(a-3\right)\left(a-1\right)\)
`4a+1(a<=0=>-a>=0)`
`=1-4(-a)`
`=1-(2sqrt{-a})^2`
`=(1-2sqrt{-a})(1+2sqrt{-a})`
đề sai rồi nha viết lại nhé