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\(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
\(1^2-2^2+3^2-....-100^2=\left(1^2-2^2\right)+...+\left(99^2-100^2\right)=\)
\(-1\left(1+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(99+100\right)=\frac{-100.101}{2}=-5050\)
\(a,49.\left(y-4\right)^2-9y^2-36y-36=49\left(y-4\right)^2-9\left(y^2+4y+4\right)\)
\(=49\left(y-4\right)^2-9\left(y+4\right)^2=\left(7y-28\right)^2-\left(3y+12\right)^2\)
\(=\left(7y-28+3y+12\right)\left(7y-28-3y-12\right)\)
\(=\left(10y-16\right)\left(4y-40\right)=8\left(5y-8\right)\left(y-10\right)\)
\(b,xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)
\(=xyz-xy-yz-xz+x+y+z-1\)
\(=\left(xyz-xy\right)-\left(xz-x\right)-\left(yz-y\right)+\left(z-1\right)\)
\(=xy\left(z-1\right)-x\left(z-1\right)-y\left(z-1\right)+\left(z-1\right)\)
\(=\left(z-1\right)\left(xy-x-y+1\right)\)
\(=\left(z-1\right)\text{[}x\left(y-1\right)-\left(y-1\right)\text{]}\)
\(=\left(z-1\right)\left(y-1\right)\left(x-1\right)\)
x^4 + 4
= x^4 + 4x^2 + 4 - 4x^2
= (x^2 + 2)^2 - 4x^2
= (x^2 + 2 - 2x)(x^2 + 2 + 2x)
(x2 - x + 1)2 - 5x(x2 - x + 1) + 4x2
Đặt x2 - x + 1 = a
<=> a2 - 5xa + 4x2 = x2 - 4xa - xa + 4x2
= a(a - 4x) - x(a - 4x) = (a - x)(a - 4x)
= (x2 - x + 1 - x)(x2 - x + 1 - 4x)
= (x2 - 2x + 1)(x2 - 5x + 1) = (x - 1)2(x2 - 5x + 1)
Đặt x2 - x + 1 = y
đthức <=> y2 - 5xy + 4x2
= y2 - xy - 4xy + 4x2
= y( y - x ) - 4x( y - x )
= ( y - x )( y - 4x )
= ( x2 - x + 1 - x )( x2 - x + 1 - 4x )
= ( x2 - 2x + 1 )( x2 - 5x + 1 )
= ( x - 1 )2( x2 - 5x + 1 )
a) (x-1)(2x+5)
b) (x+1)(x-5)
c) [(x+1)^2](x^2+x+1)
d) (x-1)(x^3-x-1)
e) (x+y)(x-y-1)
a) 2x2 + 3x - 5 = 2x2 + 5x - 2x - 5 = x(2x + 5) - (2x + 5) = (x - 1)(2x + 5)
b) x2 - 4x - 5 = x2 - 5x + x - 5 = x(x - 5) + (x - 5) = (x + 1)(x - 5)
c) x4 + x3 + x + 1 = x3(x + 1) + (x + 1) = (x + 1)(x3 + 1) = (x + 1)2(x2 - x + 1)
d) x4 - x3 - x2 + 1 = x3(x - 1) - (x - 1)(x + 1) = (x - 1)(x3 - x - 1)
e) -x - y2 + x2 - y = -(x + y) + (x - y)(x + y) = (-1 + x - y)(x + y)
Đặt \(x+y=u\)
Biểu thức trở thành \(u^2-8u+12\)
\(=u^2-2u-6u+12\)
\(=u\left(u-2\right)-6\left(u-2\right)\)
\(=\left(u-6\right)\left(u-2\right)\)
Thay ngược trở lại, ta được:
\(\left(x+y\right)^2-8\left(x+y\right)+12=\left(x+y-6\right)\left(x+y-2\right)\)
x^4 + y^4=(x^2)^2+(y^2)^2
=(x^2+y^2)^2-2x^2y^2
=(x^2+y^2)^2-(√2xy)^2
=(x^2+y^2-√2 xy)(x^2+y^2+√2 xy)