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a) x2 + 4x + 3 - y2 -2y
= x2 +4x + 4 - y2 -2y-1
= (x+2)2 - (y+1)2
= (x+2-y-1).(x+2+y+1)
= (x-y+1).(x+y+3)
b) 2a2 -5ab + 2b2
= 2a2 -4ab + 2b2 - ab
= 2.(a2 - 2ab+b2) - ab
= 2.(a-b)2 -ab
...
c) (x+y)2 - 2x - 2y + 1
= (x+y)2 - 1 - 2x -2y +2
= (x+y-1).(x+y+1) - 2.(x+y-1)
= (x+y-1)2
Ấn nhầm :v
a) \(4x^4-21x^2y^2+y^4\)
\(=\left(2x^2\right)^2-2\cdot2x^2\cdot y^2+y^2-25x^2y^2\)
\(=\left(2x^2-y^2\right)^2-\left(5xy\right)^2\)
\(=\left(2x^2-5xy-y^2\right)\left(2x^2+5xy-y^2\right)\)
b) \(x^5-5x^3+4x\)
\(=x^5-4x^3-x^3+4x\)
\(=x^3\left(x^2-4\right)-x\left(x^2-4\right)\)
\(=\left(x^2-4\right)\left(x^3-x\right)\)
\(=x\left(x-2\right)\left(x+2\right)\left(x^2-1\right)\)
\(=x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
g ) \(4x^2\left(x-2y\right)-\left(4x+1\right)\left(2y-x\right)\)
\(=4x^2\left(x-2y\right)+\left(4x+1\right)\left(x-2y\right)\)
\(=\left(4x^2+4x+1\right)\left(x-2y\right)\)
\(=\left(2x+1\right)^2\left(x-2y\right)\)
h ) \(x^2-ax^2-y+ay+cx^2-cy\)
\(=x^2\left(1-a+c\right)-y\left(1-a+c\right)\)
\(=\left(x^2-y\right)\left(1-a+c\right)\)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
a)\(x^2-y^2-2y-1=x^2-\left(y^2+2y+1\right)=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)
b)\(x^2.\left(1-x^2\right)-4+4x^2=x^2.\left(1-x^2\right)-4.\left(1-x^2\right)=\left(1-x^2\right).\left(x^2-2^2\right)\)\(=\left(1-x\right).\left(1+x\right).\left(x-2\right).\left(x+2\right)\)
Tham khảo nhé~
Bài giải:
a) x3 – 2x2 + x = x(x2 – 2x + 1) = x(x – 1)2
b) 2x2 + 4x + 2 – 2y2 = 2[(x2 + 2x + 1) – y2]
= 2[(x + 1)2 – y2]
= 2(x + 1 – y)(x + 1 + y)
c) 2xy – x2 – y2 + 16 = 16 – (x2 – 2xy + y2) = 42 – (x – y)2
= (4 – x + y)(4 + x – y)
a) \(x^3 - 2x^2 + x\) \(= x(x^2 - 2x + 1)\)
\(= x (x - 1 )^2\)
b) \(2x^2 + 4x + 2 - 2y^2\) \(= 2(x^2 + 2x + 1 - y^2)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1^2\right)-y^2\right]\)
\(= 2 (x+1-y) (x+1+y)\)
c) \(2xy - x^2 - y^2 + 16\) \(= - (x^2 - 2xy + y^2 - 4^2)\)
\(= - [(x^2 - 2xy + y^2) - 4^2]\)
\(= - [(x-y)^2 - 4^2 ]\)
\(= - (x - y - 4) (x- y + 4)\)
\(x^2\)- 4x+4-\(y^2\)+2y-1
=\(\left(x-2\right)^2\)-(y-2y+1)
=(x-2-y+1).(x-2+y-1)