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10 tháng 11 2018
câu a đề có sai số mũ ko vậy
b) \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)
\(=\dfrac{x^3\left(x+1\right)-\left(x+1\right)}{x^4+x^3+x^2+x^2+x+1}\)
\(=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+1\right)}=\dfrac{x^2-1}{x^2+1}\)
c) \(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)
\(=\dfrac{\left(x^2+3x\right)^2-1}{x^4+6x^3+9x^2-2x^2-6x+1}\)
\(=\dfrac{\left(x^2+3x-1\right)\left(x^2+3x+1\right)}{\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1}\)
\(=\dfrac{\left(x^2+3x-1\right)\left(x^2+3x+1\right)}{\left(x^2+3x-1\right)^2}=\dfrac{x^2+3x+1}{x^2-3x+1}\)
S
23 tháng 12 2016
a, \(x^4+6x^3+7x^2-6x+1\)
\(=x^4-2x^2+1+6x^3+9x^2+6x\)
\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)
\(=\left(x^2-1+3x\right)^2\)
b, \(x^4-7x^3+14x^2-7x+1\)
\(=x^4+2x^2+1+7x^3+12x^2-7x\)
\(=\left(x^2+1\right)^2-7x\left(x^2+1\right)+12^2\)
\(=\left(x^2-1+3x\right)^2\)
c, \(12x^2-11x-36\)
\(=12x^2-27x+16x-36\)
\(=3x\left(4x-9\right)+4\left(4x-9\right)\)
\(=\left(4x-9\right)\left(3x+4\right)\)
AH
Akai Haruma
Giáo viên
10 tháng 10 2018
1)
\(15x^3+29x^2-8x-12=(15x^3+30x^2)-(x^2+2x)-(6x+12)\)
\(=15x^2(x+2)-x(x+2)-6(x+2)\)
\(=(x+2)(15x^2-x-6)=(x+2)(15x^2-10x+9x-6)\)
\(=(x+2)[5x(3x-2)+3(3x-2)]\)
\(=(x+2)(3x-2)(5x+3)\)
AH
Akai Haruma
Giáo viên
10 tháng 10 2018
2)
\(x^3+4x^2-29x+24=(x^3-x^2)+(5x^2-5x)-(24x-24)\)
\(=x^2(x-1)+5x(x-1)-24(x-1)\)
\(=(x-1)(x^2+5x-24)\)
\(=(x-1)(x^2-3x+8x-24)\)
\(=(x-1)[x(x-3)+8(x-3)]=(x-1)(x-3)(x+8)\)
NT
0
NT
1
4 tháng 12 2018
\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-1}{\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-2x^2-6x+1}\)
\(=\frac{\left(x^2+3x\right)^2-1}{\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1}\)
\(=\frac{\left(x^2+3x-1\right)\left(x^2+3x+1\right)}{\left(x^2+3x-1\right)^2}=\frac{x^2+3x+1}{x^2+3x-1}\)
Tu phuong trinh da cho suy ra
(x-1)(x+1)(x+2)(x+4)=0
x=1;-1;-1;-4
\(x^4+6x^3+7x^2-6x+1=9\)
\(\Leftrightarrow x^4+6x^3+7x^2-6x-8=0\)
\(\Leftrightarrow x^4+x^3+5x^3+5x^2+2x^2+2x-8x-8=0\)
\(\Leftrightarrow\left(x^4+x^3\right)+\left(5x^3+5x^2\right)+\left(2x^2+2x\right)-\left(8x+8\right)=0\)
\(\Leftrightarrow x^3\left(x+1\right)+5x^2\left(x+1\right)+2x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+5x^2+2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^3+5x^2+2x-8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x^3+5x^2+2x-8=0\end{cases}}\)
\(x^3+5x^2+2x-8=0\)
\(\Leftrightarrow x^3-x^2+6x^2-6x+8x-8=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(6x^2-6x\right)+\left(8x-8\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+8\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+6x+8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x^2+4x+2x+8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x\left(x+4\right)+2\left(x+4\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\\left(x+4\right)\left(x+2\right)=0\end{cases}}\)
\(\left(x+4\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=-2\end{cases}}\)
Vậy \(x=\left\{-4;-2;-1;1\right\}\)