phan tich da thuc thanh nhan tu
X3_X+3X2Y+3XY2+Y3-Y
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Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
\(x^3+y^3+z^3-3xyz\) \(=\left(x+y\right)^3-3x^2y-3xy^2+z^2-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
HỌC TỐT NHA!
ta có:
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz)
Giải:
a) \(3x^2y-6xy^2\)
\(=3xy\left(x-2y\right)\)
Vậy ...
b) \(\left(2x-a\right)x^2-\left(2x-a\right)y\)
\(=\left(2x-a\right)\left(x^2-y\right)\)
\(=\left(2x-a\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)
Vậy ...
c) \(25a^2-c^2\)
\(=\left(5a-c\right)\left(5a+c\right)\)
Vậy ...
d) \(4-36x+81x^2\)
\(=2^2-2.2.9x+\left(9x\right)^2\)
\(=\left(2-9x\right)^2\)
Vậy ...
e) \(\left(x+7\right)2-\left(2x-9\right)2\)
\(=2\left[\left(x+7\right)-\left(2x-9\right)\right]\)
\(=2\left(x+7-2x+9\right)\)
\(=2\left(16-x\right)\)
Vậy ...
f) \(x^2-6x+8\)
\(=x^2-6x+9-1\)
\(=\left(x-3\right)^2-1\)
\(=\left(x-4\right)\left(x-2\right)\)
Vậy ...
\(4x^2-y^2+8\left(y-2\right)=4x^2-y^2+8y-16=4x^2-\left(y-4\right)^2=\left(2x-y+4\right)\left(2x+y-4\right)\)
x3-x+3x2y+3xy2+y3-y
=x2(x-1)+3(x2y+xy2)+y2(y-1)
=x2(x-1)+3(x2.y+y2.x)+y2(y-1)
=x2(x-1)+3{[x(x+1)+y(y+1)]}+y2(y-1)
=x2(x-1)+3.x(x+1)+3.y(y+1)+y2(y-1)
=x2(x-1)+2x2+3.x(x+1)+3.y(y+1)+y2(y-1)+2y2-2x2-2y2
=x2(x+1)+3.x(x+1)+3.y(y+1)+y2(y+1)-2x2-2y2
=(x2+3)(x+1)+(y2+3)(y+1)-2(x2+y2)
ta có : (x*3+3x*2y+3xy*2+y*3)-(x+y)
=(x+y)*3-(x+y)
=(x+y)((X+Y)*2-1)
(x+y)(x+y+1)(x+Y-1)