1. cho 1/a +1/b+1/c=0.Ch/m 1/a^3+1/b^3+1/c^3=3/abc
2. Phân tích đa thức sau thành nhân tử:
a) x^3+y^3+z^3-3xyz
b) x^3-y^3+z^3+3xyz
c) x^3-y^3-z^3-3xyz
K
Khách
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Những câu hỏi liên quan
28 tháng 8 2019
2
a
\(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^3=\left(-z\right)^3\)
\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)
\(\Rightarrow x^3+y^3+z^3=3xy\left(x+y\right)=3xyz\)
b
Đặt \(a-b=x;b-c=y;c-a=z\Rightarrow x+y+z=0\)
Ta có bài toán mới:Cho \(x+y+z=0\).Phân tích đa thức thành nhân tử:\(x^3+y^3+z^3\)
Áp dụng kết quả câu a ta được:
\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
NB
1
T
29 tháng 8 2019
#)Giải :
a) \(x+y+z=0\Leftrightarrow x+y=-z\Leftrightarrow\left(x+y\right)^3=\left(-z\right)^3\Leftrightarrow x^3+3x^2y+3xy^2+y^3=\left(-z\right)^3\)
\(\Leftrightarrow x^3+y^3+z^3=-3x^2y-3xy^2\Leftrightarrow x^3+y^3+z^3=-3xy\left(-z\right)\) hay 3xyz (đpcm)
b) \(x=\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(\Leftrightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) (Áp dụng hằng đẳng thức)
\(\Leftrightarrow x=\left[\left(b-c\right)^3+\left(c-a\right)^3\right]+\left(a-b\right)^3\)
\(=\left[\left(b-a\right)^3+\left(c-a\right)^3\right]-3\left(b-c\right)\left(c-a\right)\left[\left(b-c\right)+\left(c-a\right)\right]+\left(a-b\right)^3\)
\(=\left(b-a\right)^3-3\left(b-c\right)\left(c-a\right)\left(b-a\right)+\left(a-b\right)^3\)
\(=\left[-\left(a-b\right)^3\right]-3\left(b-c\right)\left(c-a\right)\left[-\left(a-b\right)\right]+\left(a-b\right)^3\)
\(=-\left(a-b\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)+\left(a-b\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
NB
1
SY
6 tháng 8 2017
a) (x+y)2-(x-y)2
=(x+y)(x-y)
b)(3x+1)2-(x+1)2
=[(3x+1)+(x+1)].[(3x+1)-(x+1)]
=(3x+1+x+1)(3x+1-x-1)
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
\(\left(3x+1\right)^2-\left(3x-1\right)^2\)
\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)
\(=2\cdot6x\)
\(=12x\)
_________
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=2x\cdot2y\)
\(=4xy\)
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
\(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)
\(=2x\cdot\left(x^2+3y^2\right)\)
______
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)
HQ
1
HP
2
14 tháng 6 2016
a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=2x.2\left(2x+1\right)=4x\left(2x+1\right)\)
14 tháng 6 2016
a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)
b) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy+xz+yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=\frac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2}\)
Bài 2:
a, \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(x+y\right)z-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
2a ) 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)