ak

x8 + -7x4 + -8 = 0 Reorder the terms: -8 + -7x4 + x8 = 0 Solving -8 + -7x4 + x8 = 0 Solving for variable 'x'. Factor a trinomial. (-1 + -1x4)(8 + -1x4) = 0

bn nói j vậy bạn

Ta có : x⁴ + 8x² + 7x + 8

= x⁴ - x + 8x² + 8x + 8

= x(x³ - 1) + 8(x² + x + 1)

= x(x - 1)(x² + x + 1) + 8(x² + x + 1)

= (x² - x)(x² + x + 1) + 8(x² + x + 1)

= (x² + x + 1)(x² - x + 8) 

Học tốt nhé !

a, x^2 + 5x +4

= x^2 + 1x + 4x + 4

= (x^2 + 1x) + (4x + 4)

= x ( x + 1 ) + 4 ( x + 1 )

= (x + 1) (x + 4)

b, x^2 - 6x + 5

= x^2 - 1x - 5x + 5

= (x^2 - 1x) - (5x - 5)

= x (x - 1) - 5 (x - 1)

= (x - 1) (x - 5)

c, x^2 + 7x + 12

= x^2 + 3x + 4x + 12 

= (x^2 + 3x) + (4x + 12)

= x (x + 3) + 4 (x + 3)

= (x + 3) (x + 4)

d, 2x^2 - 5x + 3

= 2^x2 - 2x - 3x + 3

= 2x (x - 1) - 3 (x - 1)

= (x-1) (2x - 3)

e, 7x  - 3x^2 - 4

= 3x + 4x - 3x^2 - 4

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

= 3x (1 - x) + 4 (x - 1)

= 3x (1-x) - 4 (1 - x)

= (1 - x) (3x - 4)

f, x^2 - 10x + 16

= x^2 - 2x - 8x + 16

= (x^2 - 2x) - (8x - 16)

= x (x - 2) - 8 (x - 2)

= (x - 2) (x - 8)

a, (x+1)(x+4)

b,(x-5)(x-1)

c,(x+3)(x+4)

d,(2x-3)(x-1)

e,(-3x+4)(x-1)

f, (x-8)(x-2)

\(2x^4+3x^3-7x^2-6x+8\)

\(=2x^4+5x^3-2x^2-8x-2x^3-5x^2+2x+8\)

\(=x\left(2x^3+5x^2-2x-8\right)-\left(2x^3+5x^2-2x-8\right)\)

\(=\left(x-1\right)\left(2x^3+5x^2-2x-8\right)\)

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

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

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

cảm ơn

Ta có:

 \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+16\right)+8\left(x^2+10x+16\right)+16\)

\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)

\(=\left(x^2+10x+16+4\right)^2=\left(x^2+10+20\right)^2\)

k nha!!

\(\text{( x + 2 ) ( x + 4 ) ( x + 6 ) ( x + 8 ) + 16}\)

\(\text{Phân tích thành nhân tử :}\)

\(\left(x^2+10x+20\right)^2\)

b: \(\left(x^2+4\right)^2-16x^2\)

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

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

c: \(x^5-x^4+x^3-x^2\)

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

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

Lời giải:

a. Bạn xem lại đề

b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

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

c.

\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)

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

(x+2)(x+4)(x+6)(x+8)+16 

=(x+2)(x+8)(x+4)(x+6)+16

=(x²+10x+16)(x²+10x+24)+16

đặt t=x²+10x+16 ta được:

t.(t+8)+16

=t²+8t+16

=(t+4)²

thay t=x²+10x+16 ta được:

(x²+10x+16)²

=[(x+2)(x+8)]²

=(x+2)²(x+8)²

vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)²(x+8)²

(x+2)(x+4)(x+6)(x+8)+16 

=(x+2)(x+8)(x+4)(x+6)+16

=(x²+10x+16)(x²+10x+24)+16

đặt t=x²+10x+16 ta được:

t.(t+8)+16

=t²+8t+16

=(t+4)²

thay t=x²+10x+16 ta được:

(x²+10x+16)²

=[(x+2)(x+8)]²

=(x+2)²(x+8)²

vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)²(x+8)²