Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)
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 \(x\in\left\{4;\dfrac{2}{3}\right\}\)
Bài 1:
b: \(=\left(x-2y\right)\left(x+2y\right)+4\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+4\right)\)
c: \(=\left(x+y-3\right)\left(x+y+3\right)\)
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)
Bài 1:
a: \(3xy^2-12x=3x\left(y^2-4\right)=3x\left(y-2\right)\left(y+2\right)\)
b: \(x^2-4y^2+4x+8y\)
\(=\left(x-2y\right)\left(x+2y\right)+4\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+4\right)\)
a: Ta có: \(x^2-6x+9-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-y-3\right)\left(x+y-3\right)\)
b: Ta có: \(x^3+4x^2+4x\)
\(=x\left(x^2+4x+4\right)\)
\(=x\left(x+2\right)^2\)
c: Ta có: \(4xy-4x^2-y^2+9\)
\(=-\left(4x^2-4xy+y^2-9\right)\)
\(=-\left(2x-y-3\right)\left(2x-y+3\right)\)
a) 4x^2 -12x = (2x)^2 - 2.2x.3 + 3^2 - 3^2
= (2x-3)^2 - 3^2
= (2x - 3 -3)(2x-3 +3)
= 2x(2x - 6)
b) x^2 - y^2 -5x +5y = (x^2 - y^2) - (5x -5y)
= (x+y)(x-y) - 5(x-y)
= (x+y - 5)(x-y)
2. 3x(x - 5) -x +5 = 0
=>3x(x - 5) - (x -5) = 0
=> (3x - 1)(x-5) = 0
=>| 3x - 1 =0 => | 3x = 1 => |x = 1/3
| x - 5 =0 | x = 5 |x= 5