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b) Ta có nhận xét này nếu a+b+c=0 thì\(a^3+b^3+c^3=3abc\) (nếu cần chứng minh thì hỏi sau nhé)
Khi đó: tử=(x-y)(y-z)(z-x)
Mẫu nó cứ thế nào ấy. Rút gọn cũng chỉ được một chút thôi, chẳng gọn lắm
a) chịu chưa nghĩ ra
d)\(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2+2xy\right)\left(x^2+y^2-z^2-2xy\right)\)
\(=\left[\left(x^2+2xy+y^2\right)-z^2\right]\left[\left(x^2-2xy+y^2\right)-z^2\right]\)
\(=\left[\left(x+y\right)^2-z^2\right]\left[\left(x-y\right)^2-z^2\right]\)
\(=\left(x+y-z\right)\left(x+y+z\right)\left(x-y-z\right)\left(x-y+z\right)\)
e)Đặt \(x^2+3x=a\)
Có: \(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
\(=\left(a+1\right)\left(a-3\right)-5\)
\(=a^2-3a+a-3-5\)
\(=a^2-2a-8\)
\(=a^2+2x-4x-8\)
\(=a\left(a+2\right)-4\left(a+2\right)\)
\(=\left(a+2\right)\left(a-4\right)\)
\(=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)
\(=\left(x^2+x+2x+2\right)\left(x^2-x+4x-4\right)\)
\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x-1\right)+4\left(x-1\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x-1\right)\left(x+4\right)\)
\(d,\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left[\left(x^2-2xy+y^2\right)-z^2\right]\left[\left(x^2+2xy+y^2\right)-z^z\right]\)
\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
\(e,\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\left(1\right)\)
\(\text{Đặt }x^2+3x+\frac{1-3}{2}=t\)
\(\text{hay }x^2+3x-2=t\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left(t+3\right)\left(t-1\right)-5\)
\(\Rightarrow t^2-t+3t-3-5\)
\(=t^2+2t-8\)
\(=t^2-2t+4t-8\)
\(=t\left(t-2\right)+4\left(t-2\right)\)
\(=\left(t-2\right)\left(t+4\right)\left(3\right)\)
\(\text{Thay (2) vào (3),ta được:}\)
\(\left(x^2+3x-2-2\right)\left(x^2+3x-2+4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x^2-x+4x-4\right)\left(x^2+x+2x+2\right)\)
\(=\left[x\left(x-1\right)+4\left(x-1\right)\right]\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)
f: \(x^2y^2+2xy+1=\left(xy+1\right)^2\)
g: \(\left(3x-2y\right)^2+2\left(3x-2y\right)+1=\left(3x-2y+1\right)^2\)
h: \(\left(x-3y\right)^2-8\left(x-3y\right)+16=\left(x-3y-4\right)^2\)
i: \(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2=4x^2\)
a)(x+y)2-(x-y)2
=(x+y-x+y)(x+y+x-y)
=2y.2x=4xy
b)(3x+1)2-(x+1)2
=(3x+1-x-1)(3x+1+x+1)
=2x.(4x+2)
=4x(2x+1)
c) x3+y3+z3-3xyz
= (x+y)3- 3xy(x+y) +z3-3xyz
=(x+y+z)( x2+2xy+y2-xz-yz+z2)-3xy(x+y+z)
=(x+y+z)(x2+y2+z2-xy-xz-yz)
Phân tích đa thức sau thành nhân tử :
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
b) \(x^3+y^3+z^3-3xyz\)
help me
Toshiro Kiyoshi30GP
Nguyễn Đình Dũng19GP
Nguyễn Huy Thắng17GP
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Đời về cơ bản là buồn... cười!!!8GP
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Ánh Dương Hoàng Vũ6GP
B= x2(y-z)+y2(z-x)+z2(x-y)
= x2(y-z)+y2z-xy2+xz2-yz2
= x2(y-z)+yz(y-z)-x(y2-z2)
= x2(y-z)+yz(y-z)-x(y-z)(y+z)
= (y-z)(x2+yz -xy -xz)
= (y-z)[x(x-y)-z(x-y)]
= (y-z)(x-y)(x-z)