toán nâng cao phần 2

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gemini_16602

Bài 1:phân tích đa thức thành nhân tử

1) x^3-9x^2+15x+25

2) x^3-4x^2-11x+30

3) 2x^4+x^3-22x^2+15x-36

4)3x^3+5x^2-14x+4

5)2x^3-x^2-3x-1


1) $x^3-9x^2+25x+25$
= $x^3+x^2-10x^2+25x+25$
= $x^{2}(x+1)-10x(x+1)+25(x+1)$
= $(x^{2}-10x+25)(x+1)$
= $(x-5)^{2}(x+1)$
2)$x^3-4x^2-11x+30$
= $x^3+3x^2-7x^2-21x+10x+30$
= $x^2(x+3)-7x(x+3)+10(x+3)$
= $(x^2-7x+10)(x+3)$
= $(x^2-2x-5x+10)(x+3)$
= $(x-5)(x-2)(x+3)$
3)$2x^4+x^3-22x^2+15x-36$
= $2x^4+2x^3-x^3-x^2-21x^2-21x+36x+36$
= $2x^3(x+1)-x^2(x+1)-21x(x+1)+36(x+1)$
= $(2x^3-x^2-21x+36)(x+1)$
 
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G

gemini_16602

Bài 1:phân tích đa thức thành nhân tử

1) x^3-9x^2+15x+25

2) x^3-4x^2-11x+30

3) 2x^4+x^3-22x^2+15x-36

4)3x^3+5x^2-14x+4

5)2x^3-x^2-3x-1

4) $3x^3+5x^2-14x+4$
= $3x^3-x^2+6x^2-2x-12x+4$
= $3x^2(x-\dfrac{1}{3})+6x(x-\dfrac{1}{3})-12(x-\dfrac{1}{3})$
= $(3x^2+6x-12)(x-\dfrac{1}{3})$
= $3(x^2+2x-4)(x-\dfrac{1}{3})$
5) $2x^3-x^2-3x-1$
= $2x^3+x^2-2x^2-x-2x-1$
= $2x^2(x+\dfrac{1}{2})-2x(x+\dfrac{1}{2})-2(x+\dfrac{1}{2})$
= $(2x^2-2x-2)(x+\dfrac{1}{2})$
= $2(x^2-x-1)(x+\dfrac{1}{2})$
 
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