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= \(5x^3-2x^2-10x^2+4x+10x-4\)
= \(5x^2\left(x-\frac{2}{5}\right)-10x\left(x-\frac{2}{5}\right)+10\left(x-\frac{2}{5}\right)\)
=\(\left(x-\frac{2}{5}\right)\left(5x^2-10x+10\right)\)
Chuc ban hoc tot!
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
\(x^3-x^2-21x+45\)
\(=\left(x^3-3x^2\right)+\left(2x^2-6x\right)+\left(-15x+45\right)\)
\(=x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)\)
\(=\left(x^2+2x-15\right)\left(x-3\right)\)
\(=\left[\left(x^2-3x\right)+\left(5x-15\right)\right]\left(x-3\right)\)
\(=\left[x\left(x-3\right)+5\left(x-3\right)\right]\left(x-3\right)\)
\(=\left(x+5\right)\left(x-3\right)^2\)
Ta có:
x^5+x+1=(x^3+x^2)+x+1=x^2(x+1) + (x+1) =(x+1)(x^2+1)
\(=x^5+x+1=x^5-x^4+x^4-x^3+x^3-x^2+x^2+x+1\)
\(=x^4\left(x-1\right)+x^3\left(x-1\right)+x^2\left(x-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right).\left(x^3-x^2+1\right)\)
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 :)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
Trả lời:
(tớ sửa lại nhé)
x4y4 + 4
= x4y4 + 4 + 4x2y2 - 4x2y2
= ( x4y4 + 4x2y2 + 4 ) - 4x2y2
= [ (x2y2)2 + 2.x2y2.2 + 22 ] - 4x2y2
= ( x2y2 + 2 )2 - (2xy)2
= ( x2y2 + 2 - 2xy )( x2y2 + 2 + 2xy )
Trả lời:
x4y4 + 4
= x4y4 + 4 + 4x2y2 - 4x2y2
= ( x4y4 + 4x2y2 + 4 ) - 4x2y2
= [ ( xy )2 + 2.xy.2 + 22 ] - ( 2xy )2
= [ ( xy )2 + 2 ]2 - ( 2xy )2
= ( x2y2 + 2 )2 - ( 2xy )2
= ( x2y2 + 2 - 2xy )( x2y2 + 2 + 2xy )