Bài 8: Phân tích đa thức thành nhân tử bằng phương pháp nhóm các hạng tử

bn nên lên hs ông Google ý chứ tháng 12 chả ai thi hsg đâu (trừ lớp 9):)))

a) \(\left(x^2-8\right)^2+36\) = \(x^4-16x^2+64+36\)

= \(\left(x^4+20x^2+100\right)-36x^2\)

= \(\left(x^2+10\right)^2-\left(6x\right)^2\)

= \(\left(x^2+10-6x\right)\left(x^2+6x+10\right)\)

b) Bài này có 2 cách; Tui sẽ làm 1 cách còn cách còn lại tui sẽ lm ở câu c

Ta có: \(x^8+x^4+1\) = \(x^8+2x^4+1-x^4\)

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

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

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

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

a)

\(\left(x^2-8\right)^2+36=x^4-16x^2+100\\ =x^4+20x^2+10^2-36x^2\)

\(=\left(x^2+10\right)^2-\left(6x\right)^2=\left(x^2+10-6x\right)\left(x^2+10+6x\right)\)

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

2 cái còn lại mik chưa nghĩ ra :D

Dễ thấy x không thể là số chẵn, do đó x lẻ

+) x = 1 thì A =1⁴ + 4 = 5, là số nguyên tố (TM)

+) x > 1, do x lẻ nên x⁴ tận cùng là 1

=> A = x⁴ + 4 tận cùng là 5 mà x⁴ + 4 > 5 nên x⁴ + 4 là hợp số, không TM đề

Vậy x = 1

x=1

\(x+2\sqrt{xy}+y-\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3\)

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

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(x-\sqrt{xy}+y\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-x+\sqrt{xy}-y\right)\)

a) \(x^3-5x^2+x-5=0\)

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

Vì \(x^2+1>0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=5\)

Vậy...

b) Mình nghĩ là sai đề nên sửa lại nhé :

\(x^4-2x^3+10x^2-20x=0\)

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

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

\(\Leftrightarrow\left(x-2\right)x\left(x^2+10\right)=0\)

Vì \(x^2+10>0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy...

a, \(x^3-5x^2+x-5=0\)

\(\Rightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-5=0\\x^2+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x^2=-1\end{matrix}\right.\)

Với mọi giá trị của \(x\in R\) ta có: \(x^2\ge0\) mà \(-1< 0\) nên \(x=5\)

Vậy \(x=5\)

Chúc bạn học tốt!!

a)

x³ - 5x² + x - 5 = 0

=> (x³ + x) - (5x² + 5) = 0
=> x(x² + 1) - 5(x² + 1) = 0

=> (x - 5)(x² + 1) =0

=> x = 5

=> x²(x² + 1) - 2x(x² + 10

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)

= \(xy\left(x+y\right)+yz\left(y+z\right)+xyz+xz\left(x+z\right)+xyz\)

= \(xy\left(x+y\right)+\left[yz\left(y+z\right)+xyz\right]+\left[xz\left(x+z\right)+xyz\right]\)

= \(xy\left(x+y\right)+yz\left(x+y+z\right)+xz\left(x+y+z\right)\)

= \(xy\left(x+y\right)+\left(x+y+z\right)\left(yz+xz\right)\)

= \(xy\left(x+y\right)+\left(x+y+z\right).z\left(x+y\right)\)

= \(xy\left(x+y\right)+\left(xz+yz+z^2\right)\left(x+y\right)\)

= \(\left(x+y\right)\left(xy+yz+xz+z^2\right)\)

= \(\left(x+y\right)[\left(xy+yz)+(xz+z^2\right)]\)

= \(\left(x+y\right)[y\left(x+z)+z(x+z\right)]\)

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

Tham khảo :

Phân tích đa thức sau thành nhân tử? | Yahoo Hỏi & Đáp

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

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

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

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

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

= \(a^2\left(b-c\right)+bc(b-c)-\left(b^{ }-c\right)\left(ab+ac\right)\)

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

= \(\left(b-c\right)\left[\left(a^2-ab\right)-\left(ac-bc\right)\right]\)

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

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

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

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

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(x^2+8x+11=t\) , ta có :

\(\left(t-4\right)\left(t+4\right)+15\)

\(=t^2-16+15=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)\)

\(=\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

Ta có:A=(x+1)(x+3)(x+5)(x+7)+15

=[(x+1)(x+7)][(x+3)(x+5)]+15

=(x²+8x+7)(x²+8x+15)+15

Đăt:t=x²+8x+7

Khi đó:A=t(t+8)+15

=>A=t²+8t+15=>A=(t+3)(t+5)

hay A=(x²+8x+10)(x²+8x+12)

\(x^3+x^2+x+1=0\)

\(\Rightarrow\left(x^3+x^2\right)+\left(x+1\right)=0\)

\(\Rightarrow x.\left(x+1\right)+\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)

Vậy \(x=-1\)

Chúc bạn học tốt!!!

Ta có:x³+x²+x+1=0

<=>(x³+x)+(x²+1)=0<=>x(x²+1)+(x²+1)=0

<=>(x+1)(x²+1)=0

=>x+1=0(vì x²+1>0)=>x=-1

Vậy S={-1}

\(x^3-x^2-x+1=0\)

\(\Rightarrow\left(x^3-x^2\right)-\left(x-1\right)=0\)

\(\Rightarrow x^2.\left(x-1\right)-\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right).\left(x^2-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x^2-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x^2=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=\pm1\end{matrix}\right.\)

Vậy \(x\in\left\{-1;1\right\}\)

Chúc bạn học tốt!!!

\(x^3-x^2-x+1=0\)

\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy x = 1 hoặc x = -1