Phân thích đa thứ thành nhân tử
49(y-4)2-9(y+2)2
8x3+1/27
125-x6
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(4\left(x+y+z\right)^2-9\left(x-y-z\right)^2\)
\(=\left[2\left(x+y+z\right)\right]^2-\left[3\left(x-y-z\right)\right]^2\)
\(=\left[2x+2y+2z\right]^2-\left[3x-3y-3z\right]^2\)
\(=\left[\left(2x+2y+2z\right)-\left(3x-3y-3z\right)\right].\left[\left(2x+2y+2z\right)+\left(3x-3y-3z\right)\right]\)
\(=\left(-x+5y+5z\right)\left(5x-y-z\right)\)
b, \(25\left(x-3y\right)^2-4\left(x+3y\right)^2\)
\(=\left[5\left(x-3y\right)\right]^2-\left[2\left(x+3y\right)\right]^2\)
\(=\left[5x-15y\right]^2-\left[2x+6y\right]^2\)
\(=\left[5x-15y-2x-6y\right].\left[5x-15y+2x+6y\right]\)
\(=\left(3x-21y\right)\left(7x-9y\right)\)
\(=3\left(x-7y\right)\left(7x-9y\right)\)
Chúc bạn học tốt.
\(=\left(3x-3y\right)^2-\left(2x+2y\right)^2=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)
\(=\left(x-5y\right)\left(5x-y\right)\)
`9(x-y)^2-4(x+y)^2`
`=[3(x-y)]^2-[2(x+y)]^2`
`=(3x-3y)^2-(2x+2y)^2`
`=(3x-3y+2x+2y)(3x-3y-2x-2y)`
`=(5x-y)(x-5y)`
\(9\left(x-y\right)^2-4\left(x+y\right)^2\\ =\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\\ =\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
Bài làm:
Ta có: \(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)
Học tốt!!!!
Ta có :
\(9\left(x-y\right)^2-4\left(x+y\right)^2=9x^2-18xy+9y^2-4x^2-8xy-4y^2\)
\(=5x^2-26xy+5y^2==\left(5x-y\right)\left(x-5y\right)\)
a, \(8^3yz+12^2yz+6xyz+yz\)
\(=512yz+144yz+6xyz+yz\)
\(=yz\left(512+14+6x+1\right)\)
\(=yz\left(527+6x\right)\)
$---$
b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)
\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)
\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)
\(=\left(z-y\right)\left(z+y\right)\left(9x^2-1\right)\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left(3x-1\right)\left(3x+1\right)\left(9x^2+1\right)\)
$---$
c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)
\(=\left[\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{3}\right)^3\right]+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}\right)+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)
$---$
d, \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2+1\right)\)
$Toru$
\(a,4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
\(b,25x^2-0,09\)
\(=\left(5x\right)^2-\left(0,3\right)^2\)
\(=\left(5x-0,3\right)\left(5x+0,3\right)\)
\(d,\left(x-y\right)^2-4\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
\(e,9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left[3-\left(x-y\right)\right]\left[3+\left(x-y\right)\right]\)
\(=\left(3-x+y\right)\left(3+x-y\right)\)
\(=\left(-x+y+3\right)\left(x-y+3\right)\)
\(f,\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2+4-4x\right)\left(x^2+4+4x\right)\)
\(=\left(x^2-2\cdot x\cdot2+2^2\right)\left(x^2+2\cdot x\cdot2+2^2\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
#\(Toru\)
a)49(y-4)2-9(y+2)2
=[7(y-4)]2-[3(y+2)]2
=(7y-28+3y+6)(7y-28-3y-6)
=(10y-22)(4y-34)
=4(5y-11)(2y-34)
b)8x3+1/27
=(2x)3+(1/3)3
=(2x+1/3)[(2x)2-2x.1/3+(1/3)2]
=(2x+1/3)(4x2-2/3x+1/9)
c)125-x6
=53-(x2)3
=(5-x2)(52+5.x2+x4)
=(5-x)(25+5x2+x4)
49(y-4)2-9(y+2)2 = [7(y-4)]2-[3(y+2)]2=[7(y-4)-3(y+2)].[7(y-4)+3(y+2)] = (7y-28-3y-6)(7y-28+3y+6) = (4y-34)(10y-22)=4(2y-17)(5y-11)