Câu hỏi của Alex Ich

Question

Phân tích đa thức thành nhân tử

a)$4x^2-12xy+9y^2$

b)$x^4+2x^2+1$

c)$a^4+4+4a$

d)$-x^2-2xy-y^2$

e)$10a-a^2-25$

g)$a^2-2a^2+1$

h)$\left(x+y\right)^2-2\left(x+y\right)+1$

lê thị hương giang · Accepted Answer

$a,4x^2-12xy+9y^2$

$=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2$

$=\left(2x-3y\right)^2$

$b,x^4+2x^2+1$

$=\left(x^2\right)^2+2.x^2.1+1^2$

$=\left(x^2+1\right)^2$

$c,a^4+4+4a$

$d,-x^2-2xy-y^2$

$=-\left(x^2+2xy+y^2\right)$

$=-\left(x+y\right)^2$

$e,10a-a^2-25$

$=-\left(a^2-10a+25\right)$

$=-\left(a^2-5a-5a+25\right)$

$=-\left[\left(a^2-5a\right)-\left(5a-25\right)\right]$

$=-\left[a\left(a-5\right)-5\left(a-5\right)\right]$

$=-a\left(a-5\right)+5\left(a-5\right)$

$=\left(a-5\right)\left(5-a\right)$

$g,a^2-2a+1$

$=a^2-2.a.1+1^2$

$=\left(a-1\right)^2$

$h,\left(x+y\right)^2-2\left(x+y\right)+1$

$=\left(x+y-1\right)^2$Câu trả lời: $a,4x^2-12xy+9y^2$