Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

ta có: \(A=8^{3^{100}}-1=\left(8^{150}\right)^2-1^2=\left(8^{150}-1\right)\left(8^{150}+1\right)\)

vậy A là hợp số.

\(8^{3^{100}}-1=8^{300}-1=\left(8^{150}\right)^2-1^2=\left(8^{150}-1\right)\left(8^{150}+1\right)\left(ĐPCM\right)\)

CM gì vậy bạn?

Áp dụng hằng đẳng thức nâng cao :
\(\left(x+y+z\right)^3=x^3+y^3+z^3+3\left(x+y+z\right)\left(xy+yz+xz\right)-3xyz\)

Vậy \(\left(x+y+z\right)^3-x^3-y^3-z^3=x^3+y^3+z^3+3\left(x+y+z\right)\left(xy+yz+xz\right)-3xyz-x^3-y^3-z^3 =3\left(x+y+z\right)\left(xy+yz+xz\right)-3xyz\)

Ta có:

\(x^8+98x^4+1\)

= \(\left(x^8+2x^4+1\right)+96x^4\)

= \(\left(x^4+1\right)^2+16\left(x^4+1\right)^2-16\left(x^4+1\right)^2+64x^4+32x^4\)

= \(\left[\left(x^4+1\right)^2+16\left(x^4+1\right)^2+64x^4\right]-\left[16\left(x^4+1\right)^2-32x^4\right]\)

= \(\left(x^4+1+8x^2\right)^2-16x^2\left(x^4+1-2x^2\right)\)

= \(\left(x^4+1+8x^2\right)^2-\left(4x\right)^2\left(x^2-1\right)^2\)

= \(\left(x^4+1+8x^2\right)^2-\left(4x^3-4x\right)^2\)

= \(\left(x^4+1+8x^2-4x^3+4x\right)\left(x^4+1+8x^2+4x^3-4x\right)\)

Ta có: \(x^4-5x^2y^2+4y^4=\left(x^4-4x^2y^2+4y^4\right)-x^2y^2\)

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

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

1,

20x⁷ + 7x - 6

= 20x⁷ - 8x + 15x - 6

= (20x⁷ - 8x )+ (15x - 6)

= 4x( 5x - 2 ) + 3( 5x - 2)

= ( 5x - 2 )( 4x + 3 )

2/

x⁴ - 5x²y² + 4y⁴

= x⁴ - 4x²y² - x²y² + 4y²

= (x⁴ - 4x²y² + 4y²) - x²y²

= ( x² - 2y²)² - x²y²

= ( x² - 2y²)² - (xy)²

= ( x² - 2y² - xy)( x² - 2y² + xy)

3/

x⁸ + y⁴ + 1

= ( x⁸ - x² ) + ( x⁴ + 1 + x² )

= x²( x⁶ - 1) + ( x⁴ + 1 + x² )

= x²( x³ - 1 )( x³ + 1) + ( x⁴ + 1 + x² )

= x² . ( x - 1 ).( x² + x + 1 ). ( x - 1).(x² - x + 1) + ( x⁴ + 1 + x² )

= x² . \(\left[\left(x-1\right)\left(x+1\right)\right].\)\(\left[\left(x^2+x+1\right)\left(x^2-x+1\right)\right]\)+ ( x⁴ + 1 + x² )

= x²(x² - 1 )\(\left[\left(x^2+1\right)^2-x^2\right]\)+ ( x⁴ + 1 + x² )

= x²(x² - 1 )( x⁴ + 2x² + 1 - x² ) + ( x⁴ + 1 + x² )

= x²(x² - 1 )( x⁴ +x² + 1 ) + ( x⁴ + 1 + x² )

= ( x⁴ - x² )( x⁴ +x² + 1 ) + ( x⁴ + 1 + x² )

= ( x⁴ +x² + 1 )( x⁴ - x² + 1 )

Nếu đề là 20x⁷ + 7x - 6 thì là đề sai, phải ko?

1) Ta có: \(4x^8+1=\left(4x^8+4x^4+1\right)-4x^4\)

= \(\left(2x^4+1\right)^2-\left(2x\right)^2\)

= \(\left(2x^4-2x^2+1\right)\left(2x^4+2x^2+1\right)\)

4) Ta có: \(\left(x^2-3x+5\right)^2-7x\left(x^2-3x+5\right)+12x^2\) (1)

Đặt \(x^2-3x+5=a\) => (1) trở thành:

\(a^2-7ax+12x^2\) = \(\left(a^2-3ax\right)-\left(4ax-12x^2\right)\)

= \(a\left(a-3x\right)-4x\left(a-3x\right)\)

= \(\left(a-3x\right)\left(a-4x\right)\) (2)

Thay a = \(x^2-3x+5\) vào 2 ta được:

\(\left(x^2-6x+5\right)\left(x^2-7x+5\right)\)

= \(\left[\left(x^2-x\right)-\left(5x-5\right)\right]\left(x^2-7x+5\right)\)

= \(\left[x\left(x-1\right)-5\left(x-1\right)\right]\left(x^2-7x+5\right)\)

= \(\left(x-5\right)\left(x-1\right)\left(x^2-7x+5\right)\)

Làm lại câu 1:

Ta có: \(4x^8+1\) = \(\left(4x^8+4x^4+1\right)-4x^4\)

= \(\left(2x^4+1\right)^2-\left(2x^2\right)^2\)

= \(\left(2x^4-2x^2+1\right)\left(2x^4+2x^2+1\right)\)

\(=2(x^2+5x-2)^2-2(x^2+5x-2)(x+3)-5(x^2+5x-2)(x+3)+2(x^2+5x-2)^2 =2(x^2+5x-2)(5x+1)-5(x^2+3)(5x+1)\)

bạn tự đặt (5x+1) ra nhé!

à mà phải là \((x^2+3)\) tớ gõ thiếu

Ta có: \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

= \(a^2\left(b-c\right)+b^2\left[\left(c-b\right)+\left(b-a\right)\right]+c^2\left(a-b\right)\)

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

= \(\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)\)

= \(\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)\)

= \(\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)\)

= \(\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

Cách này khác vs các bạn còn lại mà rất hay đc sử dụng nhiều, nhớ tick mình nha .
Giải Đặt cái trên =A
Thay a bởi b ta có A= \(a^2\left(a-c\right)+a^2\left(c-a\right)+0=a^2\left(a-c+c-a\right)=0\) Vậy A chia hết a-b . Tương tự A chia hết b-c và c-a
Vậy A = k(a-b)(b-c)(c-a) ( k là số nguyên)

Vì đẳng thức của A luôn đúng nên ko mất tính tổng quát nếu chọn a=2 b=1 c=0 ta đc 4.1 + 1(-2) + 0 = k.1.1.(-2) vậy -2 = -2k vậy k= -1
Vậy P = -(a-b)(b-c)(c-a) = (a-b)(a-c)(b-c)

Bài 2:

Cách 1:

\(x^3-7x-6=x^3-3x^2+3x^2-9x+2x-6\)

\(=\left(x^3-3x^2\right)+\left(3x^2-9x\right)+\left(2x-6\right)\)

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

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

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

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

\(=\left(x-3\right).\left[x.\left(x+1\right)+2.\left(x+1\right)\right]\)

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

Cách 2:

\(x^3-7x-6=x^3+x^2-x^2-x-6x-6\)

\(=\left(x^3+x^2\right)-\left(x^2+x\right)-\left(6x+6\right)\)

\(=x^2.\left(x+1\right)-x.\left(x+1\right)-6.\left(x+1\right)\)

\(=\left(x+1\right).\left(x^2-x-6\right)\)

\(=\left(x+1\right).\left(x^2+2x-3x-6\right)\)

\(=\left(x+1\right).\left[\left(x^2+2x\right)-\left(3x+6\right)\right]\)

\(=\left(x+1\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)

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

Chúc bạn học tốt!!! Còn 1 cách nữa nhưng mình mỏi tay quá!!!

a, \(x^3-9x^2+6x+16=x^3+x^2-10x^2-10x+16x+16\)

\(=\left(x^3+x^2\right)-\left(10x^2+10x\right)+\left(16x+16\right)\)

\(=x^2.\left(x+1\right)-10x.\left(x+1\right)+16.\left(x+1\right)\)

\(=\left(x+1\right).\left(x^2-10x+16\right)\)

\(=\left(x+1\right).\left(x^2-2x-8x+16\right)\)

\(=\left(x+1\right).\left[\left(x^2-2x\right)-\left(8x-16\right)\right]\)

\(=\left(x+1\right).\left[x.\left(x-2\right)-8.\left(x-2\right)\right]\)

\(=\left(x+1\right).\left(x-2\right).\left(x-8\right)\)

Chúc bạn học tốt!!!

b, \(x^3-x^2-x-2=x^3-2x^2+x^2-2x+x-2\)

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

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

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

Chúc bạn học tốt!!!

a, 27x³ - 27x² +18x-4=27x³ - 9x² -18x² +6x +12x-4

= 9x²(3x-1) - 6x(3x-1) + 4(3x-1)

= (3x-1)(9x²-6x+4)

b, 2x³-x²+5x+3 =\(2x^3+x^2-2x^2-x+6x+3\)

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

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

Sai đề chăng?