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a, \(9x^3y^2-15x^2y^3=3x^2y^2\cdot\left(3x-5y\right)\)
b,\(25x^2-49y^2=\left(5x\right)^2-\left(7y\right)^2\)
\(=\left(5x-7y\right)\cdot\left(5x+7y\right)\)
c,\(x^2y-xy^2-7x+7y=\left(x^2y-xy^2\right)-\left(7x-7y\right)\)
\(=xy\left(x-y\right)-7\left(x-y\right)\)
,\(=\left(x-y\right)\cdot\left(xy-7\right)\)
d, \(x^2-2xy+y^2-9z^2=\left(x^2-2xy+y^2\right)-9z^2\)
\(=\left(x-y\right)^2-9z^2\)
\(=\left(x-y+3z\right)\cdot\left(x-y-3z\right)\)
f) \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
g. \(x^{^3}+3x^2+3x+1-27z^3\\ =\left(x^{^3}+3x^2+3x+1\right)-27z^3\\ =\left(x+1\right)^3-27z^3\\ =\left(x+1-3\right)\left[\left(x+1\right)^2+\left(x+1\right)3z+9z^2\right]\\ =\left(x-2\right)\left(x+2x+1+3zx+3z+9z^2\right)\\ =\left(x-2\right)\left(3x+3zx+3z+9z^2+1\right)\left(x-2\right)3x\left(1+z\right)+3z\left(1+z\right)+1\\ =\left(x-2\right)\left(1+z\right)\left(3x+3z\right)+1\\ =\left(x-2\right)\left(1-z\right)3\left(x+z\right)+1\)
Mk lm hơi tắt, bn chú ý nha:
a,\(x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
=\(\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
=\(\left(x+1\right)^2\left(x^2-x+1\right)\)
b,\(\left(x^4-x^3\right)-\left(x^2-1\right)\)
=\(x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
= \(\left(x-1\right)\left(x^3-x-1\right)\)
c,Đề phải thế này nha:
\(x^2y-xy^2-x+y\)=\(xy\left(x-y\right)-\left(x-y\right)\)
=\(\left(x-y\right)\left(xy-1\right)\)
d,hình như đề sai đó bn, thế này đúng ko?
\(a^2x+a^2y-7x-7y\)=\(a^2\left(x+y\right)-7\left(x+y\right)\)=\(\left(x+y\right)\left(a^2-7\right)\)
e,\(4x^2-x^2-16y^2+4y^2\)
=\((4x^2-16y^2)-\left(x^2-4y^2\right)\)
=\(4\left(x-2y\right)\left(x+2y\right)-\left(x^2-2y\right)\left(x^2+2y\right)\)=\(3\left(x-2y\right)\left(x+2y\right)\)
Cách này nhanh hơn:\(3\left(x^2-4y^2\right)\)
=\(3\left(x-2y\right)\left(x+2y\right)\)
g,\(\left(x+1\right)^3-\left(3z\right)^3\)=
\(\left(x-3z+1\right)[\left(x+1\right)^2+3z\left(x+1\right)+9z^2]\)Nếu thấy đề bn đưa sai thì nhắc mk nhé?
Mong các bn giúp đỡ thêm
Chúc các bn hc tốt
\(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(\Leftrightarrow\left(x+2\right)3x\)
a) \(\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
b) \(x^2+y^2+2xy+yz+xz\)
\(=\left(x^2+2xy+y^2\right)+\left(yz+xz\right)\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
c) \(x^2-10xy-1+25y^2\)
\(=\left(x^2-10xy+25y^2\right)-1\)
\(=\left(x-5y\right)^2-1\)
\(=\left(x-5y-1\right)\left(x-5y+1\right)\)
d) \(ax^2-ax+bx^2-bx+a+b\)
\(=(ax^2+bx^2)-(ax+bx)+(a+b)\)
\(=x^2(a+b)-x(a+b)+(a+b)\)
\(=(a+b)(x^2-x+1)\)
e)\(x^2-2y+3xz+x-2y+3z\)
\(=(x^2+x)-(2xy+2y)+(3xz+3z)\)
\(=x(x+1)-2y(x-1)+3z(x+1)\)
\(=(x+1)(x-2y+3z)\)
f) \(xyz-xy-yz-xz+x+y+z-1\)
\(=(xyz-xy)-(yz-y)-(xz-x)+(z-1)\)
\(=xy(z-1)-y(z-1)-x(z-1)+(z-1)\)
\(=(z-1)(xy-y-x+1)\)
\(=(z-1)[y(x-1)-(x-1)]\)
\(=(z-1)(x-1)(y-1)\)
_Học tốt_
a,(x-y)^2-2(x+y)+1 b, x^2-y^2+4x+4 c, 4x^2-y^2+8(y-2)
=(x-y-1)^2 =(x^2+4x+4)-y^2 =4x^2-y^2+8y-16
=(x+2)^2-y^2 =4x^2-(y^2-8y+16)
=(x+2-y)(x+2+y) =4x^2-(y-4)^2
a) (x+y)2-2(x+y)+1=(x+y-1)2
b) x2-y2+4x+4 = (x2+4x+4)-y2=(x+2)2-y2=(x+y+2)(x-y+2)
c)4x2-y2+8(y-2) = 4x2-(y2-8y+16) = (2x)2-(y-4)2=(2x+y-4)(2x-y+4)
d)x3-2x2+2x-4 = x2(x-2)+2(x-2) = (x-2)(x2+2)
e)xy-4+2x-2y=x(y+2) - 2(y+2) = (x-2)(y+2)
a: \(=3x^2y^2\left(3x-5y\right)\)
b: \(=\left(5x-7y\right)\left(5x+7y\right)\)
c: \(=xy\left(x-y\right)-7\left(x-y\right)=\left(x-y\right)\left(xy-7\right)\)
d: =(x+4)(x+5)
e: \(=\left(x-y\right)^2-9z^2=\left(x-y-3z\right)\left(x-y+3z\right)\)
f: \(=\left(x^2-1\right)\left(x^2-4\right)=\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
b)\(x^2y+xy^2-x-y\)
\(=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(xy-1\right)\left(x+y\right)\)
a)\(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)