phan tich da thuc thanh nhan tu (x^2+y^2+z^2)^2- 2(x^4+y^4+z^4)
phan tich da thuc thanh nhan tu
x^2-x-y^2-y
x^2-2xy+y^2-z^2
bai 32 va 33 sbt
lop 8 bai phan tich da thuc thanh nhan tu bang cach nhom hang tu
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
con bai 32, 33 neu ban tra loi duoc minh h them
phan tich da thuc thanh nhan tu
x2(y-z) + y2(z-x) + z2(x-y)
phan tich da thuc thanh nhan tu
x^2+y^2+z^2+3xyz
Phan tich da thuc thanh nhan tu
x^2 + y^2 + z^2 - 3xyz
phan tich da thuc thanh nhan tu
x^2+y^2+z^2+3xyz
phan tich da thuc sau thanh nhan tu:
a) (3x-2)(4x-3)-(2-3x)(x-1)-2(3x-2)(x+1)
b) x^2(y-z)+y^2(z-x)+z^2(x-y)
phan tich da thuc thanh nhan tu:
a,x^4-2x^3-12x^2+12x+36
b,x^4+x^3+6x^2+5x+5
c,x^8y^8+x^4y^4+1
d,x^5-x^4+x^3-X^2+x-1
e,x^5+x^4-X63+x62-x+2
g,x(Y^2-z^2)+y(z^2-x^2)+z(x^2-y^2)
1 Phan tich da thuc thanh nhan tu
b) 4x(x+y)(x+y+z)(x+z) +y2z2 ( phân tích thành số chính phương)
4x(x+y)(x+y+z)(x+z)+(yz)^2
=(2x(x+y+z))(2(x+y)(x+z)+(yz)^2
=(2x^2+2xy+2xz)(2x^2+2xy+2xz+2yz)+(yz)^2
Đặt t=C
=(t-yz)(t+yz)-(yz)^2
=t^2-(yz)^2+(yz)^2=t^2=(2x^2+2xy+2xz+yz)^2
=
cam on ban nhung mk lam xong roi
phan tich da thuc sau thanh nhan tu :
\(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3
= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3
Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0
=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3
= 3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )
= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )