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a)\(x^2+y^2-x^2y^2+xy-x-y\)
\(=-\left(x-1\right)\left(y-1\right)\left(x+y+xy\right)\)
b)\(x^2(1-x^2)-4-4x^2\)
\(=-\left(x^2-x+2\right)\left(x^2+x+2\right)\)
c)\((1+2x)(1-2x)-x(x+2)(x-2)\)
\(=-\left(x-1\right)\left(x^2+5x+1\right)\)
\(5x^2-8xy-4y^2\)
\(=\left(5x^2-10xy\right)+\left(2xy-4y^2\right)\)
\(=5x\left(x-2y\right)+2y\left(x-2y\right)\)
\(=\left(5x+2y\right)\left(x-2y\right)\)
a) = (3x)^2 + 2.3x.5+ 5^2 = (3x+5)^2
b) = (2/3x)^2-(4y)^2=(2/3x-4y)(2/3x+4y)
c) = -(9x^4-12/5x^2y^2+4/25y^4) = -[(3x^2)^2 - 2.3x^2.2/5y^2 + (2/5y^2)^2]= -(3x^2-2/5y^2)^2
d) = (x-5)^2 - 4^2= (x-5+4)(x-5-4) = (x-1)(x-9)
e) = (2x)^3 + 3.(2x)^2.(5y) + 3.(2x).(5y)^2 + (5y)^3 = (2x+5y)^3
f) = (8x)^2 - (8a+b)^2 = (8x-8a-b)(8x+8a+b)
g) = (7x-4-2x-1)(7x-4+2x+1) = (5x-5)(9x-3) = 5(x-1).3(x-3)=15(x-1)(x-3)
h) = (x-y)(x+y)- 2(x+y) = (x+y)(x-y-2)
# Chúc bạn học tốt #
a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz
= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]
= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)
= (xy + yz + zx)(x + y + z)
b) Vô câu hỏi tương tự
a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz
= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]
= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)
= (xy + yz + zx)(x + y + z)
b) tương tự
a)3xy+x+15y+5
=(3xy+x)+(15y+5)
=x(3y+1)+5(3y+1)
=(x+y)(3y+1)
b)2x+2y+x2-y2
=tôi đang nghĩ
a, 3xy+x+15y+5
=x(3y+1)+5(3y+1)
=(3y+1)(x+5)
b, 2x+2y+x2-y2
=2(x+y)+(x-y)(x+y)
=(x+y)(2+x-y)
bài 1:
\(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
\(\Rightarrow x=0\)(vì \(x^2+1\ne0\))
bài 2:
\(x\left(x+y\right)-5x-5y\)
\(=x\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x-5\right)\left(x+y\right)\)
***lesson 1
x^3 - x = 0
x(x^2 - 1) = 0
Since anything time zero is zero, either one of the terms in the multiplication must be zero.
First multiplicand is x^2 - 1.
x^2 - 1 = 0
x^2 = 1
x = 1 and -1
Second multiplicand is x. This answer is merely:
x = 0
So the answers to your question are x = -1, 0 and 1.