Question

[Đăng Văn Nghĩa]

Phân tích thành nhân tử dùng phương pháp hằng đẳng thức;
a)\(\frac{1}{4}\left(a+b\right)^2-1\)

b) \(9\left(x-y\right)^2-4\left(x-y\right)^2\)

c) \(\left(p-2q\right)^2-4\left(p+q\right)^2\)

d) \(25p^2m^4-\frac{1}{36}p^4\)

Answer 1

[Phan Nghĩa]

a,\(\frac{1}{4}\left(a+b\right)^2-1=\left(\frac{a+b}{2}\right)^2-1^2=\left(\frac{a+b}{2}-1\right)\left(\frac{a+b}{2}+1\right)\)

b,\(9\left(x-y\right)^2-4\left(x-y\right)^2=\left(x-y\right)^2\left(9-4\right)=5\left(x-y\right)^2\)

c,\(\left(p-2q\right)^2-4\left(p+q\right)^2=\left(p-2q-2p-2q\right)\left(p-2q+2p+2q\right)\)

\(=\left(-p-4q\right)3p\)

d, \(25p^2m^4-\frac{1}{36}p^4=\left(5pm^2\right)^2-\left(\frac{p^2}{6}\right)^2=\left(5pm-\frac{p^2}{6}\right)\left(5pm+\frac{p^2}{6}\right)\)

Answer 2

[Nguyễn Huy Tú]

a, \(\frac{1}{4}\left(a+b\right)^2-1=\left(\frac{1}{2}a+\frac{1}{2}b\right)^2-1=\left(\frac{a}{2}+\frac{b}{2}-1\right)\left(\frac{a}{2}+\frac{b}{2}+1\right)\)

b, \(9\left(x-y\right)^2-4\left(x-y\right)^2=\left(3x-3y\right)^2-\left(2x-2y\right)^2\)

\(=\left(3x-3y-2x+2y\right)\left(3x-3y+2x-2y\right)=5\left(x-y\right)^2\)

c, \(\left(p-2q\right)^2-4\left(p+q\right)^2=\left(p-2q\right)^2-\left(2p+2q\right)^2\)

\(=\left(p-2q-2p-2q\right)\left(p-2q+2p+2q\right)^2=9p^2\left(-p-4q\right)\)

d, \(25p^2m^4-\frac{1}{36}p^4=\left(5pm^2\right)^2-\left(\frac{1}{6}p^2\right)^2=\left(5pm^2-\frac{1}{6}p^2\right)\left(5pm^2+\frac{1}{6}p^2\right)\)

\(=p^2\left(5m^2-\frac{1}{6}p\right)\left(5m^2+\frac{1}{6}p\right)\)