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\(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(\Leftrightarrow\left(x+2\right)3x\)
a)\(\left(4x^3-xy^2+y^3\right)\left(x^2y+2xy^2-2y^3\right)\)
\(=x^2y\left(4x^3-xy^2+y^3\right)+2xy^2\left(4x^3-xy^2+y^3\right)\)
\(-2y^3\left(4x^3-xy^2+y^3\right)\)
\(=4x^5y-x^3y^3+x^2y^4+8x^4y^2-2x^2y^4+2xy^5\)
\(-8x^3y^3+2xy^5-2y^6\)
\(=-2y^6+4x^5y+\left(2xy^5+2xy^5\right)+8x^4y^2+\left(x^2y^4-2x^2y^4\right)\)
\(-\left(x^3y^3+8x^3y^3\right)\)
\(=-2y^6+4x^5y+4xy^5+8x^4y^2-x^2y^4-9x^3y^3\)
b)
(!) \(2\left(x+y\right)^2-7\left(x+y\right)+5\)
\(=2\left(x+y\right)^2-2\left(x+y\right)-5\left(x+y\right)+5\)
\(=2\left(x+y\right)\left(x+y-1\right)-5\left(x+y-1\right)\)
\(=\left(2x+2y-5\right)\left(x+y-1\right)\)
(!!) \(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-x^2-y^2-z^2\)
\(=2\left(xy+yz+zx\right)\)
a, \(=12x^5+9x^3y^2-6x^2y^3-20x^4y-15x^2y^3-10xy^4-24x^3y^2-18xy^4+12y^5\)
(tự rút gọn cái :P)
b, \(8x^3+4x^2y-2xy^2-y^3\)
\(=4x^2\left(2x+y\right)-y^2\left(2x+y\right)=\left(2x+y\right)^2\left(2x-y\right)\)
\(4x^2y^2-4x^2-4xy-y^2=4x^2y^2-\left(2x+y\right)^2\)
\(=\left(2x+y+2xy\right)\left(2xy-2x+y\right)\)
Mấy cái còn lại nhân tung ra là được mà :))))
1, x2+3xy+2y2= x2+xy+2xy+2y2=x(x+y)+2y(x+y)=(x+2y)(x+y)
2, x(x+2)(x+3)(x+5)+9=x(x+5)(x+2)(x+3)+9=(x2+5x)(x2+5x+6)+9
Đặt x2+5x=t, ta có
t(t+6)+9=t2+6t+9=(t+3)2=(x2+5x+3)2=(x2+8)2
3, x2+2xy+y2+2x+2y-15=(x+y)2+2(x+y)-15=(x+y)2+2(x+y)+1-16=(x+y+1)2-42
= (x+y+1-4)(x+y+1+4)=(x+y-3)(x+y+5)
4, 4x4y4+1=4x4y4+4x2y2+1-4x2y2=(2x2y2+1)2-(2xy)2=(2x2y2+1-2xy)(2x2y2+1+2xy)
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_
\(a,\left(2x+1\right)^2-\left(x-1\right)^2\\ =\left(2x+1-x+1\right)\left(2x+1+x-1\right)\\ =\left(x+2\right)3x\)
\(b,9\left(x+5\right)^2-\left(x-7\right)^2\\ =\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2\\ =\left(3x+15-x+7\right)\left(3x+15+x-7\right)\\ =\left(2x+22\right)\left(4x+8\right)\)
\(c,x^2y+xy^2-x-y\\ =xy\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(xy-1\right)\)
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