Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) x2 + 4x – y2 + 4;
=x2+4x+4-y2
=(x+2)2-y2
=(x+2-y)(x+2+y)
b) 3x2 + 6xy + 3y2 – 3z2;
=3.(x2+2xy+y2)-3z2
=3.(x+y)2-3z2
=3.[(x+y)2-z2]
=3.(x+y-x)(x+y+z)
c) x2 – 2xy + y2 – z2 + 2zt – t2.
=(x-y)2-(z2-2zt+t2)
=(x-y)2-(z-t)2
=[(x-y)-(z-t)][(x-y)+(z-t)]
=(x-y-z+t)(x-y+z-t)
a) Ta có : x2 + 4x – y2 + 4
= x2 + 4x + 4 - y2
= (x + 2)2 - y2
= (x + 2 - y)(x + 2 + y)
b) 3x2 + 6xy + 3y2 - 3z2
= 3(x2 + 2xy + y2) - 3z2
= 3(x + y)2 - 3z2
= 3[(x + y)2 - z2]
= 3(x + y - z)(x + y + z)
\(x^2+4x-y^2+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x-y+2\right)\left(x+y+2\right)\)
hk tốt
^^
\(x^2-3x+xy-3y\)
\(=\left(x^2+xy\right)-\left(3x+3y\right)\)
\(=x.\left(x+y\right)-3.\left(x+y\right)\)
\(=\left(x-3\right).\left(x+y\right)\)
\(2x^2-x+2xy-y\)
\(=2x^2-\left(x-2xy+y\right)\)
\(=2x^2-\left(x-y\right)^2\)
\(=\left(\sqrt{2}x\right)^2-\left(x-y\right)^2\)
\(=\left(\sqrt{2}x-x+y\right).\left(\sqrt{2}x+x-y\right)\)
\(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x.\left(x^2+1\right)\)
\(=\left(x^2+1\right).\left(x^2+1+x\right)\)
\(16+2xy-x^2-y^2\)
\(=16-x^2+2xy-y^2\)
\(=16-\left(x^2-2xy+y^2\right)\)
\(=4^2-\left(x-y\right)^2\)
\(=[4-\left(x-y\right)].[4+\left(x-y\right)]\)
\(=\left(4-x+y\right).\left(4+x-y\right)\)
a) x2 + 4x – y2 + 4;
=x2+4x+4-y2
=(x+2)2-y2
=(x+2-y)(x+2+y)
b) 3x2 + 6xy + 3y2 – 3z2;
=3.(x2+2xy+y2)-3z2
=3.(x+y)2-3z2
=3.[(x+y)2-z2]
=3.(x+y-x)(x+y+z)
c) x2 – 2xy + y2 – z2 + 2zt – t2.
=(x-y)2-(z2-2zt+t2)
=(x-y)2-(z-t)2
=[(x-y)-(z-t)][(x-y)+(z-t)]
=(x-y-z+t)(x-y+z-t)
a; \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2=\left(x+y-2\right)\left(x-y+2\right)\)
b; \(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)
c, \(x^2-2xy+y^2-z^2+2zt-t^2=\left(x-y\right)^2-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t^2\right)=\left(x-y-z+t\right)\left(x+y+z-t\right)\)
a) \(9x^2-12x+4\)
\(=9x^2-6x-6x+4\)
\(=3x\left(3x-2\right)-2\left(3x-2\right)\)
\(=\left(3x-2\right)^2\)
b) \(2xy+16-x^2-y^2\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left(x-y\right)^2+16\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)
c) \(3x+2x^2-2\)
\(=2x^2+4x-x-2\)
\(=2x\left(x+2\right)-\left(x+2\right)=\left(x+2\right)\left(2x-1\right)\)
\(a,x^2+6x+9\)
\(=x^2+3x+3x+9\)
\(=\left(x^2+3x\right)+\left(3x+9\right)\)
\(=x.\left(x+3\right)+3.\left(x+3\right)\)
\(=\left(x+3\right).\left(x+3\right)\)
\(=\left(x+3\right)^2\)
\(b,10x-25-x^2\)
\(=-\left(x^2-2.5.x+5^2\right)\)
\(=-\left(x-5\right)^2\)
\(c,x^2+4x-y^2+4\)
\(=\left(x^2+2.2.x+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right).\left(x+2+y\right)\)
\(d,3x^2+6xy+3y^2-3z^2\)
\(=3.[\left(x^2+2xy+y^2\right)-z^2]\)
\(=3.[\left(x+y\right)^2-z^2]\)
\(=3.\left(x+y-z\right)\left(x+y+z\right)\)
\(e,x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=[\left(x-y\right)-\left(z-t\right)].[\left(x-y\right)+\left(z-t\right)]\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Bài 1: 4a2-4ab+b2-9a2b2
=(2a)2-2.2a.b+b2-(3ab)2
=(2a-b)2-(3ab)2
=(2a-b-3ab)(2a-b+3ab)
a/ (4a2-4ab+b2)-9a2b2
= (2a-b)2-(3ab)2
= (2a-b-3ab) (2a-b+3ab)
a) \(x^2+4x-y^2+4=\left(x^2+4x+4\right)-y^2\)
\(\left(x+2\right)^2-y^2=\left(x+2-y\right).\left(x+2+y\right)\)
b) \(3x^2+6xy+3y^2-3z^2\Leftrightarrow\left(\sqrt{3}x+\sqrt{3}y\right)^2-\left(\sqrt{3}z\right)^2\)
\(\Leftrightarrow\left(\sqrt{3}x+\sqrt{3}y-\sqrt{3}z\right).\left(\sqrt{3}x+\sqrt{3}y+\sqrt{3}z\right)\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\Leftrightarrow\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(\Leftrightarrow\left(x-y\right)^2-\left(z-t\right)^2=\left(x-y-z+t\right)\left(x-y+z-t\right)\)