phân tích đa thức thành nhân tử
x^4-2x^3+2x-1
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\(x^4+2x^3+10x-25\)
\(=x^4+5x^2+2x^3+10x-5x^2-25\)
\(=\left(x^2+5\right)\left(x^2+2x-5\right)\)
\(x^4+x^3+2x^2+x+1=\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+1\right)\left(x^2+x+1\right)\)
Dễ thấy \(x^2+1>0\); \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) nên ta không thể phân tích thêm được nữa.
Vậy \(x^4+x^3+2x^2+x+1=\left(x^2+1\right)\left(x^2+x+1\right)\).
Bài 2:
Sửa đề: \(x^3-3x^2-10x=0\)
\(\Leftrightarrow x\left(x^2-3x-10\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-2\end{matrix}\right.\)
Câu 1:
\(=x^2-\left(y-4\right)^2\)
\(=\left(x-y+4\right)\cdot\left(x+y-4\right)\)
a: \(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)\)
\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
b: \(x^3-x^2-x-2\)
\(=x^3-2x^2+x^2-2x+x-2\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x-2\right)\cdot\left(x^2+x+1\right)\)
c: \(x^3+x^2-x+2\)
\(=x^3+2x^2-x^2-2x+x+2\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-x+1\right)\)
d: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
e: Sửa đề: \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
f: \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
g: \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
h: \(\left(x^2-3\right)^2+16\)
\(=x^4-6x^2+9+16\)
\(=x^4-6x^2+25\)
\(=x^4+10x^2+25-16x^2\)
\(=\left(x^2+5\right)^2-\left(4x\right)^2\)
\(=\left(x^2+5+4x\right)\left(x^2+5-4x\right)\)
Lời giải:
$=x+\sqrt{x}(\sqrt{y}+\sqrt{2})-\sqrt{3}(\sqrt{y}+\sqrt{2})-3$
$=(x-3)+\sqrt{x}(\sqrt{y}+\sqrt{2})-\sqrt{3}(\sqrt{y}+\sqrt{2})$
$=(\sqrt{x}-\sqrt{3})(\sqrt{x}+\sqrt{3})+(\sqrt{y}+\sqrt{2})(\sqrt{x}-\sqrt{3})$
$=(\sqrt{x}-\sqrt{3})(\sqrt{x}+\sqrt{3}+\sqrt{y}+\sqrt{2})$
Câu này số xấu không phân tích được nhân tử lun
a: \(=x^2\left(x-2\right)\)
b: \(=\left(y+1\right)^2-x^2=\left(y+1+x\right)\left(y+1-x\right)\)
c: =(x-3)(x+2)
x^4-2x^3+2x-1
=x4-1-2x3+2x
=(x2-1)(x2+1)-2x(x2-1)
=(x2-1)(x2-2x+1)
=(x-1)(x+1)2(x-1)2