a, bằng cách tìm nhân tử chung
1,\(x^2-3x\)
=x.(\(\left(x-3\right)\)
2,\(15x^2-6x\)
=3x.(5x-2)
3,\(4x\left(x-y\right)\)\(+2y\left(x-y\right)\)
=(x-y).(4x+2y)
=2(x-y).(x+y)
=2(\(x^2-y^2\left(\right)\)
b, dùng hằng đẳng thức
1,\(64x^2-25y^2\)
=\(\left(8x\right)^2-\left(5y\right)^2\)
=(8x-5y)(8x+5y)
2,\(9x^2-30x-25\)
=\(\left(3x-5\right)^2\)
3,
\(\dfrac{1}{4}x^2+2x+4\)
=\(\left(\dfrac{1}{2}x+2\right)^2\)
4,\(25a^2-2a+\dfrac{1}{25}\)
=(\(\left(5a-\dfrac{1}{5}\right)^2\)
Phân tích hằng đẳng thức
a) Bằng cách tìm nhân tử chung
1) \(x^2-3x=x\left(x-3\right)\)
2) \(15x^2-6x=3x\left(5x-2\right)\)
3) \(4x\left(x-y\right)+2y\left(x-y\right)=\left(4x+2y\right)\left(x-y\right)\)
4) \(5\left(a-2b\right)-10x\left(b-2a\right)=5a-10b-10xb+20xa=5\left(a-2b-2xb+4xa\right)\)
b) Bằng cách dùng hằng đẳng thức
1) \(64x^2-25y^2=\left(8x-5y\right)\left(8x+5y\right)\)
2) \(9x^2-30x+25=\left(3x\right)^2-2.3x.5+5^2=\left(3x-5\right)^2\)
3) \(\dfrac{1}{4}x^2+x+4=\left(\dfrac{1}{2}x\right)^2+\dfrac{1}{2}.x.2+2^2=\left(\dfrac{1}{2}x+2\right)^2\)
4) \(25a^2-2a+\dfrac{1}{25}=\left(5a\right)^2-5.2.a+\left(\dfrac{1}{5}\right)^2=\left(5a-\dfrac{1}{5}\right)^2\)
Phân tích đa thức thành nhân tử
a) Bằng cách tìm nhân tử chung
1) \(x^2-3x=x\left(x-3\right)\)
2) \(15x^2-6x=3x\left(5x-2\right)\)
3) \(4x\left(x-y\right)+2y\left(x-y\right)=\left(4x+2y\right)\left(x-y\right)\)
4) \(5\left(a-2b\right)-10x\left(b-2a\right)=5a-10b-10xb+20xa=5\left(a-2b-2xb+4xa\right)\)
b) Bằng hằng đẳng thức
1) \(64x^2-25y^2=\left(8x+5y\right)\left(8x-5y\right)\)
2) \(9x^2-30x+25=\left(3x\right)^2-2.3x.5+5^2=\left(3x-5\right)^2\)
3) \(\dfrac{1}{4}x^2+2x+4=\left(\dfrac{1}{2}x\right)^2+2.\dfrac{1}{2}x.2+2^2=\left(\dfrac{1}{2}x+2\right)^2\)
4) \(25a^2-2a+\dfrac{1}{25}=\left(5a\right)^2-2.5a\dfrac{1}{5}+\left(\dfrac{1}{5}\right)^2=\left(5a-\dfrac{1}{5}\right)^2\)