Bài 6: Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung

Ta có: \(81^7-27^9-9^{13}\)\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)

\(=3^{28}-3^{27}-3^{26}\)

\(=3^{22}.\left(3^6-3^5-3^4\right)\)

\(=3^{22}.\left(729-243-81\right)\)

\(=3^{22}.405⋮405\)

Vậy \(81^7-27^9-9^{13}⋮405\)

\(81^7-27^9-9^{13}\)

\(=3^{28}-3^{27}-3^{26}\)

\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)

\(=\left(3^{26}.3^2\right)-\left(3^{26}.3\right)-\left(3^{26}.1\right)\)

\(=3^{26}\left(3^2-3-1\right)\)

\(=3^{26}.5\)

\(=3^{22}\left(2^3.5\right)\)

\(=3^{22}.405⋮405\)

\(\Leftrightarrow81^7-27^9-9^{13}⋮405\)

\(\rightarrowđpcm\)

\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)

\(=3^{28}-3^{27}-3^{26}\)

\(=\left(3^{26}.3^2\right)-\left(3^{26}.3\right)-3^{26}\)

\(=3^{26}\left(3^2-3-1\right)\)

\(=3^{26}.5\)

\(=3^{22}.3^3.5\)

\(=3^{22}.405⋮405\)

\(\Leftrightarrow81^7-27^9-9^{13}⋮405\rightarrowđpcm\)

\(a\left(a+2b\right)^3-b\left(2a+b\right)^3=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^2\right)\)

\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)

\(=a^4-2a^3b+2ab^3-b^4\)

\(=\left(a^4-b^4\right)-\left(2a^3b-2ab^3\right)\)

\(=\left(a^2-b^2\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2-2ab\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(a-b\right)^2=\left(a-b\right)^3\left(a+b\right)\)

\(x^3-4x^2+2x^2-1\)

= \(x^3-2x^2+2x-1\)

= \(x^3-x^2-x^2+x+x-1\)

= \(x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)

= \(\left(x-1\right)\left(x^2-x+1\right)\)

Ta thấy \(x^2-x+1=\left(x-\dfrac{1}{4}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Vậy x=1

14x²-21xy²+28x²y²

=7x(2x-3xy²+4xy²)

Phân tích đa thức thành nhân tử.

14x² -21xy²+28x²y²

= 7x(2x-3y²+4xy²)

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

\(=x^2\left(x^3-3x^2+3x-1\right)\)

\(=x^2\left(x^3-3.x^2.1+3.x.1^2-1^3\right)\)

\(=x^2\left(x-1\right)^3\)

b)3y^2- 3z^2+ 3x^2+ 6xy

\(=3\left(y^2-z^2+x^2+2xy\right)\)

\(=3\left(\left(x^2+2.x.y+y^2\right)-z^2\right)\)

\(=3\left(\left(x+y\right)^2-z^2\right)\)

\(=3\left(x+y+z\right)\left(x+y-z\right)\)

c)x^2- 6x +5

\(=x^2-6x+9-4\)

\(=\left(x^2-2.x.3+3^2\right)-2^2\)

\(=\left(x-3\right)^2-2^2\)

=(x-3-2)(x-3+2)

=(x-5)(x-1)

d)x^2 + 5x -6

\(=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{49}{2}\)

\(=\left(x+\dfrac{5}{2}\right)^{2 }-\left(\dfrac{7}{2}\right)^2\)

\(=\left(x+\dfrac{5}{2}+\dfrac{7}{2}\right)\left(x+\dfrac{5}{2}-\dfrac{7}{2}\right)\)

=(x+6)(x-1)

e)x^2 - 5x + 4

=\(x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{9}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\)

\(=\left(x-\dfrac{5}{2}-\dfrac{3}{2}\right)\left(x-\dfrac{5}{2}+\dfrac{3}{2}\right)\)

=(x-4)(x-1)

Gọi x là số phải tìm.

Theo đề bài, ta có: 3x²=2x². Giải phương trình ta có x=0

Vậy số cần tìm là 0.