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a: \(=x\left(x-3\right)-4y\left(x-3\right)\)
=(x-3)(x-4y)
d: \(=\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left(x+2\right)\left(x-2+x+2\right)\)
=2x(x+2)
\(a,=x\left(x-3\right)-4y\left(x-3\right)=\left(x-4y\right)\left(x-3\right)\\ b,=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)=\left(x-1\right)\left(x^2-3x+1\right)\\ c,=\left(x-y\right)\left(1-a\right)\\ d,=\left(x-2\right)\left(x-2+x+2\right)=2x\left(x-2\right)\\ e,=x^2\left(x+y\right)-xz\left(x+y\right)=x\left(x-z\right)\left(x+y\right)\\ f,=\left(x-y-2\right)\left(x+y\right)\)
\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)
\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)
\(c.x^2-4xy+4y^2-4y\\ =\left(x-2y\right)^2-4y\\ =\left(x-2y\right)^2-\left(2y\right)^2\\ =\left(x-2y+2y\right)\left(x-2y-2y\right)\\ =x\left(x-4y\right)\)
\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)
b: =xy-x-y+1
=x(y-1)-(y-1)
=(x-1)(y-1)
c: =(x-2y)^2-4y
\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)
d: =16-(x^2-2xy+y^2)
=16-(x-y)^2
=(4-x+y)(4+x-y)
a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
\(a.3x^2-3y^2-2\left(x-y\right)^2\\ =3\left(x^2-y^2\right)-2\left(x-y\right)^2\\ =3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\\ =\left(x-y\right)\left[3\left(x+y\right)-2.\left(x-y\right)\right]=\left(x-y\right)\left(x+5y\right)\\ b.x^2-y^2-2x-2y\\ =\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\\ =\left(x+y\right)\left(x-y-2\right)\\ c.\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\\ =\left(x-1\right)\left(2x+1\right)\left[1+3\left(x+2\right)\right]\\ =\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\\ d.\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\\ =\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\\ =\left(x-5\right)\left[\left(x-5\right)+\left(x+5\right)+\left(2x+1\right)\right]\\ =\left(x-5\right)\left(4x+1\right)\)
a: \(x^4+4=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b: \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
c: \(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
\(a,x\left(x+6\right)\\ b,\left(9x-1\right)\left(9x+1\right)\\ c,\left(x+y\right)-3^2\\ =\left(x+y-3\right)\left(x+y+3\right)\\ d,\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x+y-1\right)\)
a) \(40x^4-10x^2=10x^2\left(4x^2-1\right)=10x^2\left(2x-1\right)\left(2x+1\right)\)
b) \(16x^4-20x^2-y^2-5y=\left(4x^2-\dfrac{5}{2}\right)^2-\left(y-\dfrac{5}{2}\right)^2=\left(4x^2-\dfrac{5}{2}-y+\dfrac{5}{2}\right)\left(4x^2-\dfrac{5}{2}+y-\dfrac{5}{2}\right)=\left(4x^2-y\right)\left(4x^2+y-5\right)\)c)\(64a^2-9b^2-16a+1=\left(8a-1\right)^2-9b^2=\left(8a-1-3b\right)\left(8a-1+3b\right)\)d) \(5x^2+23x-10=5\left(x-\dfrac{2}{5}\right)\left(x+5\right)\)
a: \(40x^4-10x^2\)
\(=10x^2\left(4x^2-1\right)\)
\(=10x^2\cdot\left(2x-1\right)\left(2x+1\right)\)
b: \(16x^4-20x^2-y^2-5y\)
\(=\left(4x^2-y\right)\left(4x^2+y\right)-5\left(4x^2+y\right)\)
\(=\left(4x^2+y\right)\left(4x^2-y-5\right)\)
c: Ta có: \(64a^2-9b^2-16a+1\)
\(=\left(8a-1\right)^2-9b^2\)
\(=\left(8a-1-3b\right)\left(8a-1+3b\right)\)
d: Ta có: \(5x^2+23x-10\)
\(=5x^2+25x-2x-10\)
\(=\left(x+5\right)\left(5x-2\right)\)
a) \(2x^2-50\)
\(=2\left(x^2-25\right)\)
\(=2\left(x-5\right)\left(x+5\right)\)
b) \(x^2z+4xyz+4y^2z\)
\(=z\left(x^2+4xy+4y^2\right)\)
\(=z\left(x+2y\right)^2\)
c) \(x^2-y^2+12y-36\)
\(=x^2-\left(y^2-2\cdot x\cdot6+6^2\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right)\left(x+y-6\right)\)
d) Đặt \(D=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(D=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(D=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(x^2+3x+1=a\)
\(D=\left(a-1\right)\left(a+1\right)+1\)
\(D=a^2-1^2+1\)
\(D=a^2\)
Thay \(x^2+3x+1=a\)vào D ta có :
\(D=\left(x^2+3x+1\right)^2\)
a: \(2x^2-50=2\left(x^2-5^2\right)\)
\(=2\left(x-5\right)\left(x+5\right)\)
b:\(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)\)
\(=z\left(x+2y\right)^2\)
c:\(x^2-y^2+12y-36=x^2-\left(y^2-12y+36\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right)\left(x+y-6\right)\)
d:\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(t=x^2+3x+1\)Ta có:
\(=\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1=t^2\)
\(=\left(x^2+3x+1\right)^2\)