PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ
n7+ n2 + 1
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Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(=mp\left(m^2+mn-mp-np\right)=mp\left[m\left(m+n\right)-p\left(m+n\right)\right]=mp\left(m+n\right)\left(m-p\right)\)
b) \(=abm^2+abn^2+a^2mn+b^2mn=am\left(bm+an\right)+bn\left(bm+an\right)\)
\(=\left(bm+an\right)\left(am+bn\right)\)
\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)
\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)
\(=3\left(a-b+c\right)\left(x+6y\right)^2\)
___________________
\(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
\(a,=xy\left(-6x+y\right)\)
\(b,=10c\left(a^2-9b^2+3bc-ac\right)=10c\left[\left(a-3b\right)\left(a+3b\right)-c\left(a-3b\right)\right]\)
\(=10c\left[\left(a-3b\right)\left(a+3b-c\right)\right]\)
c,\(=a\left(x-c\right)-b\left(x-c\right)=\left(a-b\right)\left(x-c\right)\)
d,\(=-\left(x-2y-6\right)\left(x-2y+6\right)\)
e;\(=m^2+4m+mn+n^2+4n+mn=m\left(m+4+n\right)+n\left(m+4+n\right)\)\(=\left(m+n\right)\left(m+n+4\right)\)
f,\(=\dfrac{1}{2}\left(4x^2-y^2\right)=\dfrac{1}{2}\left(2x-y\right)\left(2x+y\right)\)
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
a: \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)
\(=x^2y^2+a^2b^2+x^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(b^2+y^2\right)\left(x^2+a^2\right)\)
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
\(a,Sửa:16-9x^2+y^2-8y\\ =\left(y-4\right)^2-9x^2\\ =\left(y-3x-4\right)\left(y+3x-4\right)\\ b,=\left(n^2+4n\right)^2+10\left(n^2+4n\right)+25\\ =\left(n^2+4n+5\right)^2\)
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
n^9 + 1 = 1n^9
\(n^7+n^2+1\)
\(=\left(n^7-n^6+n^4-n^3+n^2\right)\)\(+\left(n^6-n^5+n^3-n^2+n\right)\)
\(+\left(n^5-n^4+n^2-n+1\right)\)
\(=n^2\left(n^5-n^4+n^2-n+1\right)\)\(+n\left(n^5-n^4+n^2-n+1\right)\)
\(+\left(n^5-n^4+n^2-n+1\right)\)
\(=\left(n^2+n+1\right)\left(n^5-n^4+n^2-n+1\right)\)