K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TD
24 tháng 5 2017
a ) \(\left(x+y+z\right)^2=x^2+y^2+z^{2^{ }}+2xy+2yz+2zx\)
Biến đổi vế trái ta được :
\(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)\)
\(=x^2+xy+xz+xy+y^2+yz+zx+zy+z^2\)
\(=x^2+y^2+z^{2^{ }}+2xy+2yz+2zx\)
Vậy \(\left(x+y+z\right)^2=x^2+y^2+z^{2^{ }}+2xy+2yz+2zx\)
HL
0
Y
1 tháng 8 2018
1) \(a^4-3a^3-6a^3+18a^2-18a^2+54a+27a-81\)
\(=a^3\left(a-3\right)-6a^2\left(a-3\right)-18a\left(a-3\right)+27\left(a-3\right)\)
\(=\left(a-3\right)\left(a^3-6a^2-18a+27\right)\)
\(=\left(a-3\right)\left(a^3+3a^2-9a^2-27a+9a+27\right)\)
\(=\left(a-3\right)\left[a^2\left(a+3\right)-9a\left(a+3\right)+9\left(a+3\right)\right]\)
\(=\left(a-3\right)\left(a+3\right)\left(a^2-9a+9\right)\)
2) Ta có:
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y+z\right)-x\right]\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz+x^2+xy+xz+x^2\right)-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right)\left(3x^2+3xy+3xz+2yz+y^2+z^2\right)-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right)\left(3x^2+3xy+3xz+2yz+y^2+z^2-y^2+yz-z^2\right)\)
\(=\left(y+z\right)\left(3x^2+3xy+3xz+3yz\right)\)
\(=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
NT
1
PL
10 tháng 6 2018
a, x^4 - 5x^2 + 4
= x^4 - 4x^2- x+ 4
= x^2 . (x^2 - 4) - (x^2 - 4)
= (x^2 - 4) . (x^2 - 1)
= (x - 2) . (x + 2) . (x - 1) . (x + 1)
TN
2
7 tháng 7 2017
\(\left(x+y+z\right)^3-x^3-y^3-z^3\\ =x^3+y^3+z^3-x^3-y^3-z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\\ =3\left(x+y\right)\left(y+z\right)\left(z+x\right)\:\left(đpcm\right)\)
Q
7 tháng 7 2017
• \(VT=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3z\left(x+y\right)^2+3\left(x+y\right)z^2+z^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3xy^2+y^3+3z+\left(x+y\right)^2+3xz^2+3yz^2-x^3-y^3\)
\(=3x^2y+3xy^2+3z\left(x^2+2xy+y^2\right)+3xz^2+3yz^2\)
\(=3x^2y+3xy^2+3x^2z+6xyz+3y^2z+3xz^2+3yz^2\) (1)
• \(VP=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(=\left(3x+3y\right)\left(y+z\right)\left(z+x\right)\)
\(=\left(3xy+3xz+3y^2+3yz\right)\left(z+x\right)\)
\(=3xyz+3x^2y+3xz^2+3x^2z+3y^2z+3xy^2+3yz^2+3xyz\)
\(=6xyz+3x^2y+3xz^2+3x^2z+3y^2z+3xy^2+3yz^2\) (2)
Từ (1) và (2) suy ra \(VT=VP\) (đpcm)
NT
1
6 tháng 7 2016
\(\left(x+y+z\right)^3=\left[\left(x+y\right)+z\right]^3=\left(x+y\right)^3+z^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+3xy\left(x+y\right)+c^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(xy+zx+zy+z^2\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\Rightarrow\left(dpcm\right)\)
Chúc bạn học tốt
T I C K nha cảm ơn bạn
CC
1
CL
20 tháng 6 2015
x^3 + y^3 + z^3 +3(x+y)(y+z)(z+x)=x3+y3+z3+(3x+3y)(y+z)(z+x)
=x3+y3+z3+(3xy+3xz+3y2+3yz)(z+x)
=x3+y3+z3+3xyz+3x2y+3xz2+3x2z+3y2z+3y2x+3yz2+3xyz
=x3+y3+z3+3x2y+3xz2+3x2z+3y2z+3y2x+3yz2+6xyz
=x3+3x2y+3y2x+y3+3x2z+6xyz+3y2z+3xz2+3yz2+z3
=(x+y)3+3z(x2+2xy+y2)+3z2(x+y)+z3
=(x+y)3+3z(x+y)2+3z2(x+y)+z3
=(x+y+z)3
vậy (x+y+z)^3= x^3 + y^3 + z^3 +3(x+y)(y+z)(z+x)
HA
4
2 tháng 9 2017
\(VT=\left(x+y+z\right)^3=\left[\left(x+y\right)+z\right]^3\)
\(=\left(x+y\right)^3+z^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
\(=VP\left(đpcm\right)\)
2 tháng 9 2017
\(\left(x+y+z\right)^3=x^3+y^3+z^3+3x^2y+3xy^2+3y^2z+3z^2x+3x^2z+3z^2x+6xyz\)
=\(x^3+y^3+z^3+3\left(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+2xyz\right)\)
=\(x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)(đpcm)
KS
1
T
29 tháng 7 2019
#)Giải :
\(\left(x+y+z\right)^3-x^3y^3z^3=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Leftrightarrow\left(x+y+z\right)^3=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
Ta có : \(\left(x+y+z\right)^3=\left[\left(x+y\right)+z\right]^3=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(xy+yz+zx+z^2\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Rightarrow\left(x+y+z\right)^3=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Rightarrow\left(x+y+z\right)^3-x^3-y^3-z^3=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(đpcm\right)\)