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a,x^2-x-y^2-y
=x^2-y^2-(x+y)
=(x-y).(x+y)-(x+y)
=(x+y).(x-y-1)
b, x^2-2xy+y^2-z^2
=(x^2-2xy+y^2)-z^2
=(x-y)^2-z^2
=(x-y-z)(x-y+z)
c,5x-5y+ax-ay( đề bài ở đây phải là -ay ms tính đc)
=(5x-5y)+(ax-ay)
=5(x-y)+a(x-y)
=(x-y).(5+a)
d,a^3-a^2.x-ay+xy
=(a^3-a^2x)-(ay-xy)
=a^2(a-x)-y(a-x)
=(a-x)(a^2-y)
e,4x^2-y^2+4x+1
={(2x)^2+4x+1}-y^2
=(2x+1)^2-y^2
=(2x+1+y^2)(2x+1-y^2)
f,x^3-x+y^3-y
=(x^3+y^3)-(x+y)
=(x+y)(x^2-xy+y^2)-(x+y)
=(x+y)(x^2-xy+y^2-1)
a: =x(4x^2+4x+1)
=x(2x+1)^2
b: =(x-y)^2-49
=(x-y-7)(x-y+7)
2xy - 4x + 5y - 10
= (2xy - 4x) + (5y - 10)
= 2x(y - 2) + 5(y - 2)
= (y - 2)(2x + 5)
a: \(15x^2-5x^3=5x^2\left(3-x\right)\)
b: \(8x^3-y^3+4x^2y-2xy^2\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+2xy\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+4xy+y^2\right)\)
\(=\left(2x-y\right)\left(2x+y\right)^2\)
c: Ta có: \(x^8+64y^4\)
\(=x^8+16x^4y^2+64y^4-16x^4y^2\)
\(=\left(x^4+8y^2\right)^2-\left(4x^2y\right)^2\)
\(=\left(x^2-4x^2y+8y^2\right)\left(x^2+4x^2y+8y^2\right)\)
c: \(x^2-4+3\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(3x-6\right)\)
\(=\left(x-2\right)\left(x+2+3x-6\right)\)
\(=\left(4x-4\right)\left(x-2\right)\)
\(=4\left(x-1\right)\left(x-2\right)\)
a Đề sai: )
b
\(a^3-a^2x-ay+xy\\ =a^2\left(a-x\right)-y\left(a-x\right)\\ =\left(a-x\right)\left(a^2-y\right)\)
c
\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)
d
\(x^4+2x^3+x^2\\ =x^4+x^3+x^3+x^2\\ =x^3\left(x+1\right)+x^2\left(x+1\right)\\ =\left(x^3+x^2\right)\left(x+1\right)\)
e
\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)
c: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
d: =x^2(x^2+2x+1)
=x^2(x+1)^2
e: =5(x^2-2xy+y^2-z^2)
=5[(x-y)^2-z^2]
=5(x-y-z)(x-y+z)
1.
a) \(2x^4-4x^3+2x^2\)
\(=2x^2\left(x^2-2x+1\right)\)
\(=2x^2\left(x-1\right)^2\)
b) \(2x^2-2xy+5x-5y\)
\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)
\(=2x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(2x+5\right)\)
2 .
a,
\(4x\left(x-3\right)-x+3=0\)
⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)
⇒\(\left(x-3\right)\left(4x-1\right)=0\)
⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)
b,
\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)
⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0
⇒\(\left(x-4\right)\left(3x-2\right)=0\)
⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)
a: \(15a^2x-10ax^2\)
\(=5ax\cdot3a-5ax\cdot2x\)
\(=5ax\left(3a-2x\right)\)
b: \(2xy-4x+5y-10\)
\(=\left(2xy-4x\right)+\left(5y-10\right)\)
\(=2x\left(y-2\right)+5\left(y-2\right)\)
\(=\left(y-2\right)\left(2x+5\right)\)