\(3,\\ a,=x^2\left(x-1\right)+2\left(x-1\right)=\left(x^2+2\right)\left(x-1\right)\\ b,=a\left(a-b\right)-\left(a-b\right)=\left(a-1\right)\left(a-b\right)\\ c,=y\left(x^2+z\right)-5\left(x^2+z\right)=\left(y-5\right)\left(x^2+z\right)\\ d,=2y\left(x-2\right)-\left(x-2\right)=\left(2y-1\right)\left(x-2\right)\\ e,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ f,Sửa:5x^2-5xy+x-y=5x\left(x-y\right)+\left(x-y\right)=\left(5x+1\right)\left(x-y\right)\\ 4,\\ a,\Rightarrow x\left(2x-13\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{13}{2}\end{matrix}\right.\\ b,\Rightarrow-x^2+x+2+x^2-2=0\Rightarrow x=0\\ c,\Rightarrow\left(x-3\right)\left(3x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\\ d,\Rightarrow x\left(x^2+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Rightarrow x=0\)
Bài 3:
a) \(=x^2\left(x-1\right)+2\left(x-1\right)=\left(x-1\right)\left(x^2+2\right)\)
b) \(=a\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(a-1\right)\)
c) \(=y\left(x^2+z\right)-5\left(x^2+z\right)=\left(x^2+z\right)\left(y-5\right)\)
d) \(=x\left(2y-1\right)-2\left(2y-1\right)=\left(2y-1\right)\left(x-2\right)\)
e) \(=a^2\left(a-x\right)-y\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)
f) \(=5x\left(x-y\right)-\left(x-y\right)=\left(x-y\right)\left(5x-1\right)\)
Bài 4:
a) \(\Rightarrow x\left(2x-13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{13}{2}\end{matrix}\right.\)
b) \(\Rightarrow-x^2+x+2+x^2-2=0\)
\(\Rightarrow x=0\)
c) \(\Rightarrow\left(x-3\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(\Rightarrow x\left(x^2+1\right)=0\Rightarrow x=0\left(do.x^2+1\ge1>0\right)\)
Bài 3
a) x3 - x2 + 2x - 2
= (x3 - x2) + (2x - 2)
= x2(x - 1) + 2(x - 1)
= (x - 1)(x2 + 2)
b) a2 - ab - a + b
= (a2 - ab) - (a - b)
= a(a - b) - (a - b)
= (a - b)(a - 1)
c) x2y + yz - 5(x2 + z)
= (x2y + yz) - 5(x2 + z)
= y(x2 + z) - 5(x2 + z)
= (x2 + z)(y - 5)
d) 2xy + 2 - x - 4y
= (2xy - 4y) + (2 - x)
= 2y(x - 2) - (x - 2)
= (x - 2)(2y - 1)
e) a3 - a2x - ay + xy
= (a3 - a2x) - (ay - xy)
= a2(a - x) - y(a - x)
= (a - x)(a2 - y)
f) Sửa đề lại thành 5x2 - 5xy + x - y
= (5x2 - 5xy) + (x - y)
= 5x(x - y) + (x - y)
= (x - y)(5x - 1)
Bài 4:
a) 2x2 - 13x = 0
x(2x - 13) = 0
x = 0 hoặc 2x - 13 = 0
*) \(2x-13=0\)
\(x=\dfrac{13}{2}\)
Vậy \(x=0;x=\dfrac{13}{2}\)
b) (x + 1)(2 - x) + x2 - 2 = 0
2x - x2 + 2 - x + x2 - 2 = 0
x = 0
Vậy x = 0
c) 3x(x - 3) - x + 3 = 0
3x(x - 3) - (x - 3) = 0
(x - 3)(3x - 1) = 0
x - 3 = 0 hoặc 3x - 1 = 0
*) x - 3 = 0
x = 3
*) 3x - 1 = 0
\(x=\dfrac{1}{3}\)
Vậy \(x=3;x=\dfrac{1}{3}\)
d) \(x^3+x=0\)
\(x\left(x^2+1\right)=0\)
x = 0 hoặc x2 + 1 = 0 (vô lý)
Vậy x = 0
