Question

phân tích đa thức sau thành nhân tử 

e) 25 - x² + 4xy - 4y²

f) x⁴ - 4x³ + 4x²

g) x³ - x² - x + 1 

h) a⁵ + 27a²

i) x³ + 3x² - 3x - 1

k) 4a²b² - ( a² + b² - 1)²

giúp mình với giải chi tiết giúp mình nha

Answer 1

\(25-x^2+4xy-4y^2=5^2-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)

\(x^4-4x^3+4x^2=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\)

\(x^3-x^2-x+1=x^2\left(x-1\right)-\left(x-1\right)=\left(x^2-1\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)^2\)

\(a^5+27a^2=a^2\left(a^3+27\right)=a^2\left(a+3\right)\left(a^2-3a+9\right)\)

\(x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+3x\right)=\left(x-1\right)\left(x^2+4x+1\right)\)

\(4a^2b^2-\left(a^2+b^2-1\right)^2=\left(2ab+a^2+b^2-1\right)\left(2ab-a^2-b^2+1\right)=\left[\left(a+b\right)^2-1\right]\left[1-\left(a-b\right)^2\right]\)

\(\left(a+b-1\right)\left(a+b+1\right)\left(1+a-b\right)\left(1-a+b\right)\)

Answer 2

e)25-x²+4xy-4y²

=25-(x²-4xy+4y²)

=5²-(x-y)²

​=(5+x-y)(5-x+y)

 

 

Answer 3

e) \(25-x^2+4xy-4y^2=5^2-\left[x^2+4xy-\left(2y\right)^2\right]=5^2-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)

f) \(x^4-4x^3+4x^2=x^2\left(x^2-4x+4\right)=x^2\left(x^2-4x+2^2\right)=x^2\left(x-2\right)^2\)

g) \(x^3-x^2-x+1=\left(x^3-x^2\right)-\left(x-1\right)=x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)\)

h) \(a^5+27a^2=a^2\left(a^3+27\right)=a^2\left(a^3+3^3\right)=a^2\left(a+3\right)\left(a^2-3a+9\right)\)

i) \(x^3+3x^2-3x-1=\left(x^3-1\right)+\left(3x^2-3x\right)=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+3x\right)=\left(x-1\right)\left(x^2+4x+1\right)\)k) \(4a^2b^2-\left(a^2+b^2-1\right)^2=\left(2ab\right)^2-\left(a^2+b^2-1\right)^2=\left(2ab-a^2-b^2+1\right)\left(2ab+a^2+b^2-1\right)=\left[-\left(a+b\right)^2+1\right]\left[\left(a+b\right)^2-1\right]\)

\(=\left[1-\left(a+b\right)^2\right]\left[\left(a+b\right)^2-1\right]=\left(1-a-b\right)\left(1+a+b\right)\left(a+b-1\right)\left(a+b+1\right)=\left(a+b+1\right)^2\left(1-a-b\right)\left(a+b-1\right)\)