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\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y-x+y\right)\left(x+y+x-y\right)\)
\(=2y.2x\)
\(=4xy\)
b) (3x+1)2-(x+1)2
=(3x+1+x+1)[(3x+1)-(x+1)]
=(4x+2)2x
=2(2x+1)2x
=4x(2x+1)
c) x3+y3+z3-3xyz
=(x+y)3-3xy(x+y)+z3-3xyz
=[(x+y)3+z3]-3xy(x+y+z)
=(x+y+z)[(x+y)2-z(x+y)+z2]-3xy(x+y+z)
=(x+y+z)(x2+2xy+y2-xz-zy+z2-3xy)
=(x+y+z)(x2+y2+z2-xy-yz-zx)
\(\left(x^2+xy\right)^2-\left(y^2+xy\right)^2\)
\(=\left(x^2+xy-y^2-xy\right)\left(x^2+xy+y^2+xy\right)\)
\(=\left(x^2-y^2\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x+y\right)^2\)
\(=\left(x-y\right)\left(x+y\right)^3\)
•x3+y3+z3-3xyz=(x+y)3-3xy(x+y)+z3-3xyz
=(x+y+z)[(x+y)2-(x+y).z+z2]-3xy(x+y+z)
=(x+y+z)(x2+y2+z2+2xy-xz-yz) -3xy(x+y+z)
=(x+y+z)(x2+y2+z2-xy-yz-xz)
•(x2+xy)2-(y2+xy)2=[x(x+y)]2-[y(x+y)]2
=x2.(x+y)2-y2.(x+y)2
=(x+y)2.(x2-y2)=(x+y)2.(x+y).(x-y)
=(x+y)3(x-y)
•3x2-3x-36=3.(x2-x-12)
=3(x2-4x+3x-12)
=3[x(x-4)+3(x-4)]=3(x-4)(x+3)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
a) Đề bài phải là : \(\left(x+y\right)^2-\left(x-y\right)^2\)thì mới phân tích được.
Nếu đề bài như trên ta có:
\(\left(x+y\right)^2-\left(x-y\right)^2=\)\(\left(x+y-x+y\right)\left(x+y+x-y\right)=2x\cdot2y=4xy\)
b) Ta có: \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
= \(2x\cdot\left(4x+2\right)=2x\cdot2\cdot\left(2x+1\right)=4x\cdot\left(2x+1\right)\)
c) Ta có : \(x^3+y^3+z^3-3xyz\)
= \(\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xy\)
=\(\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3xy\left(x+y+z\right)\)
=\(\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
=\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
a) \([(x-y)3 + (y-z)3]+ (z-x)3\)=\(\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]=\left(x-z\right)\left[\left(x-2y+z\right)\left(x+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)=\left(x-z\right)\left(x-y\right)\left(z-y\right)3\)
b) \(=y^2\left(x^2y-x^3+z^3-z^2y\right)-z^2x^2\left(z-x\right)=y^2\left[-y\left(z^2-x^2\right)-\left(z^3-x^3\right)\right]-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(-yz-xy-z^2-zx-x^2\right)-z^2x^2\left(z-x\right)=\left(z-x\right)\left(-y^3z-xy^2-z^2y^2-xyz-x^2y^2-z^2x^2\right)\)
đến đây coi như là thành nhân tử rồi nha. em muốn gọn thì ráng ngồi nghĩ rồi tách nha. chỉ cần nhóm mấy cái có ngoặc giống nhau là đc. k khó đâu. chịu khó nghĩ để rèn luyện nha
c) \(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(\left(9a^3-6a^2\right)+\left(6a^2-4a\right)+\left(-9a+6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
d) em sửa đề đi. đề sai rồi. đồng nhất hệ số phải có dấu bằng nha.
có gì liên hệ chị. đúng nha ;)
a,81-(x^2-4xy+4y^2)=81-(x-2y)^2=(9-(x-2y))(9+(x-2y))=(9-x+2y)(9+x-2y)
b,x^3+y^3+z^3-3xyz=(x^3+3(x^2)y+3x(y^2)+y^3)+z^3-3xyz-3xy(x+y)
=((x+y)^3+3((x+y)^2)z+3(x+y)z^2+z^3)-(3xyz-3xy(x+y))-3(x+y)z(x+y+z)
=(x+y+z)^3-3(x+y)z(x+y+z)-3xy(x+y+z)=(x+y+z)((x+y+z)^2-3(x+y)z-3xy)
=(x+y+z)(x^2+y^2+z^2+2xy+2yz+2xz-3xy-3yz-3xz)=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Bài 1:
\(a,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
\(=2x\left(4x+2\right)=4x\left(2x+1\right)\)
b, \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-yz-xz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(87^2+73^2-27^2-13^2=\left(87^2-13^2\right)+\left(73^2-27^2\right)=74.100+46.100=100\cdot\left(74+46\right)=12000\)