phân tích đa thức thành nhân tử.
x^5 + x +1
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1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(x^5-x\\ =x\left(x^4-1\right)\\ =x\left(x^2-1\right)\left(x^2+1\right)\\ =x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
Ta có: \(x^5-x\)
\(=x\left(x^4-1\right)\)
\(=x\left(x^2-1\right)\left(x^2+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
\(x^2\left(x-5\right)+x-5=x^2\left(x-5\right)+\left(x-5\right)=\left(x-5\right)\left(x^2+1\right)\)
Lời giải:
$x^5+x-1=(x^5+x^2)-(x^2-x+1)$
$=x^2(x^3+1)-(x^2-x+1)=x^2(x+1)(x^2-x+1)-(x^2-x+1)$
$=(x^2-x+1)[x^2(x+1)-1]=(x^2-x+1)(x^3+x^2-1)$
x5+x+1=x3+x2+x+1=x2(x+1)+(x+1)=(x+1)(x2+1)
Với dạng phân tích đa thức bậc cao này ta thường tách như sau:
\(x^5+x+1=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
CHÚC EM HỌC TỐT :)