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1 tháng 8 2016

sarangheyo 

10 tháng 5 2022

\(x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0\)

\(\Leftrightarrow x^6-x^5-5x^5+5x^4+10x^4-10x^3-10x^3+10x^2+5x^2-5x-x+1=0\)

\(\Leftrightarrow x^5\left(x-1\right)-5x^4\left(x-1\right)+10x^3\left(x-1\right)-10x^2\left(x-1\right)+5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^5-5x^4+10x^3-10x^2+5x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^5-x^4-4x^4+4x^3+6x^3-6x^2-4x^2+4x+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^4\left(x-1\right)-4x^3\left(x-1\right)+6x^2\left(x-1\right)-4x\left(x-1\right)+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[x^4-4x^3+6x^2-4x+1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[x^4-x^3-3x^3+3x^2+3x^2-3x-x+1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^3\left[x^3-3x^2+3x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^3\left[x^3-x^2-2x^2+2x+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^4\left[x^2-2x+1\right]=0\Leftrightarrow\left(x-1\right)^6=0\Leftrightarrow x=1\)

21 tháng 7 2017

a) \(x^3+4x^2-29x+24=x^3-x^2+5x^2-5x-24x+24\)

\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+5x-24\right)\)

\(=\left(x-1\right)\left(x^2+8x-3x-24\right)\)

\(=\left(x-1\right)\left[x\left(x+8\right)-3\left(x+8\right)\right]\)

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

b) \(x^4+6x^3+7x^2-6x+1\)

\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)

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

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

c) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2=x^4-2x^3+6x^2-8x+8\)

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

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

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

d) Phức tạp mà dài quá :v

\(6x^5+15x^4+20x^3+15x^2+6x+1\)

\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)

\(=3x^4\left(2x+1\right)+6x^3\left(2x+1\right)+7x^2\left(2x+1\right)+4x\left(2x+1\right)+\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3x^4+6x^3+7x^2+4x+1\right)\)

\(=\left(2x+1\right)\left[\left(3x^4+3x^3+x^2\right)+\left(3x^3+3x^2+x\right)+\left(3x^2+3x+1\right)\right]\)

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

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

e)

- Câu này có thể áp dụng định lý: nếu tổng các hệ số biến bậc chẵn và tổng các hệ số biến bậc lẻ bằng nhau thì đa thức có nhân tử x + 1.

- Nhận thấy: 1 + 4 + 4 + 1 = 3 + 4 + 3

\(x^6+3x^5+4x^4+4x^3+4x^2+3x+1\)

\(=(x^6+x^5)+(2x^5+2x^4)+(2x^4+2x^3)+(2x^3+2x^2)+(2x^2+2x)+(x+1)\)

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

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

Tiếp tục phân tích bằng cách trên vì 1 + 2 + 2 = 2 + 2 +1

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

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

11 tháng 10 2017

a) Gọi CT ghi hóa trị của NH3\(N^xH^I_3\) (x: nguyên, dương)

Theo quy tắc hóa trị, ta có:

\(x.1=I.3\\ =>x=\dfrac{1.I}{3}=III\)

Vậy: Hóa trị của N có hóa trị III trong hợp chất NH3

b) Gọi CT kèm hóa trị của Zn(OH)2\(Zn^x\left(OH\right)^y_2\) (x,y: nguyên, dương).

Theo quy tắc hóa trị, ta có:

\(x.1=y.2\\ =>\dfrac{x}{y}=\dfrac{2}{1}=\dfrac{II}{I}\)

=> x=II

y=I

=> Hóa trị của Zn là II trong hợp chất trên

3 tháng 1 2017

a)\(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)

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

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

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

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

b)\(6x^5+15x^4+20x^3+15x^2+6x+1\)

\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)

\(=3x^4\left(2x+1\right)+6x^3\left(2x+1\right)+7x^2\left(2x+1\right)+4x\left(2x+1\right)+\left(2x+1\right)\)

\(=\left(3x^4+6x^3+7x^2+4x+1\right)\left(2x+1\right)\)

\(=\left[3x^4+3x^3+x^2+3x^3+3x^2+x+3x^2+3x+1\right]\left(2x+1\right)\)

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

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

NV
16 tháng 3 2019

a/ \(\left(x^2-x+2\right)^2+\left(x-2\right)^2=\left(x^2-x+2\right)^2-x^2+x^2+\left(x-2\right)^2\)

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

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

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

b/ \(6x^5+15x^4+20x^3+15x^2+6x+1\)

\(=6x^5+6x^4+2x^3+9x^4+9x^3+3x^2+9x^3+9x^2+3x+3x^2+3x+1\)

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

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

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

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

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

10 tháng 2 2018

a, = [(x-2).(x+1)]^2+(x-2)^2

    = (x-2)^2.(x+1)^2+(x-2)^2

    = (x-2)^2.[(x+1)^2+1]

    = (x-2)^2.(x^2+2x+2)

Tk mk nha

10 tháng 2 2018

b)  \(6x^5+15x^4+20x^3+15x^2+6x+1\)

\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)

\(=\left(2x+1\right)\left(3x^4+6x^3+7x^2+4x+1\right)\)

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

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

7 tháng 4 2019

\(6x^5+15x^4+20x^3+15x^2+6x+1 \)

\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)

\(=\left(2x+1\right)\left(3x^4+6x^3+7x^2+4x+1\right)\)

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

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

11 tháng 2 2018

a) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)

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

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

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

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

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

b) \(6x^5+15x^4+20x^3+15x^2+6x+1\)

\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)

\(=3x^4\left(2x+1\right)+6x^3\left(2x+1\right)+7x^2\left(2x+1\right)+4x\left(2x+1\right)+2x+1\)

\(=\left(2x+1\right)\left(4x^4+6x^3+7x^2+4x+1\right)\)

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

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

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