Phân tích đa thức sau thành nhân tử:
a3 -3a2 +3a-1-b2
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\(=a^3b-a^3c+b^3c-ab^3+c^3a-bc^3\)
\(=\left(a^3b-ab^3\right)+\left(a^3c-ac^3\right)+\left(b^3c-bc^3\right)\)
\(=ab\left(a^2-b^2\right)+ac\left(a^2-c^2\right)+bc\left(b^2-c^2\right)\)
Bài 1:
a: \(4a^2-6b=2\left(2a^2-3b\right)\)
b: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
Bài 1:
a) \(4a^2-6b=2\left(a^2-3b\right)\)
b) \(=mn\left(m^2-2mn-1\right)\)
Bài 2:
a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)
b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)
a2 – b2 – 4a + 4
= a2 – 4a + 4 – b2
= (a – 2)2 – b2
= (a – 2 + b)(a – 2 – b)
= (a + b – 2)(a – b – 2)
\(4a^3-3a+1\)
\(=\left(4a^3-4a\right)+\left(a+1\right)\)
\(=4a\left(a^2-1\right)+\left(a+1\right)\)
\(=4a\left(a-1\right)\left(a+1\right)+\left(a+1\right)\)
\(=\left(a+1\right)\left(4a^2-4a+1\right)\)
\(=\left(a+1\right)\left(2a-1\right)^2\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
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: \(3abc^3-6a^2b^3c+12a^3bc\)
\(=3abc\cdot c^2-3abc\cdot2ab^2+3abc\cdot4a^2\)
\(=3abc\left(c^2-2ab^2+4a^2\right)\)
b: \(27-8y^3\)
\(=3^3-\left(2y\right)^3\)
\(=\left(3-2y\right)\left(9+6y+4y^2\right)\)
c: Sửa đề: \(4x^2+4x-y^2+1\)
\(=\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+1+y\right)\left(2x+1-y\right)\)
d: \(3a^2\cdot\left(x-2\right)-6ab\cdot\left(2-x\right)\)
\(=3a^2\cdot\left(x-2\right)+6ab\cdot\left(x-2\right)\)
\(=\left(x-2\right)\left(3a^2+6ab\right)\)
\(=3a\left(a+2b\right)\left(x-2\right)\)
a) (a + b)2 – m2 + a + b – m = (a + b + m)(a + b – m) + (a + b – m)
= (a + b – m)(a + b + m + 1)
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) \(=abm^2+abn^2+a^2mn+b^2mn=am\left(bm+an\right)+bn\left(bm+an\right)\)
\(=\left(bm+an\right)\left(am+bn\right)\)