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a) -5x2+x+15x-3 = \(-5x\left(x-\frac{1}{5}\right)+15\left(x-\frac{1}{5}\right)\)=(3-x)(5x-1)

b)x2+x-6x-6 = x(x+1)-6(x+1) = (x-6)(x+1)

c) x2-x-6x+6 = x(x-1)-6(x-1) = (x-6)(x-1)

3 tháng 7 2018

cj xem lại đề rồi trả lời lại hộ e e cần gấp,e cảm ơn

15 tháng 10 2021

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(5x-1\right)\)

c:\(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)

\(=\left(2x-1\right)\left(-3x+2\right)\)

15 tháng 10 2021

1.

a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

2.

a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

28 tháng 10 2017

a, \(x^4+2x^2+1-x^2\)

\(\left(x^2+1\right)^2-x^2\)

\(\left(x^2+x+1\right)\left(x^2-x+1\right)\)

b, \(x^4+x^2+1\)

\(x^4+2x^2+1-x^2\)

= .. ( như phần a )

c, \(y^4+64\)

\(\left(y^2+8\right)\left(y^2-8\right)\)

d, \(4xy+3z-12y-xz\)

\(=4y\left(x-3\right)-z\left(x-3\right)\)

\(=\left(x-3\right)\left(4y-z\right)\)

e, \(x^2-4xy+4y^2-z^2+6z-9\)

\(=\left(x-2y\right)^2-\left(z-3\right)^2\)

g, \(x^2-4xy+5x+4y^2-10y\)

\(=\left(x^2-4xy+4y^2\right)+\left(5x-10y\right)\)

\(=\left(x-2y\right)^2+5\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-2y+5\right)\)

h, \(x^2-7x+6\)

\(=x^2-6x-x+6\)

\(=x\left(x-6\right)-\left(x-6\right)\)

\(=\left(x-6\right)\left(x-1\right)\)

i, \(x^3+5x^2+6x+2\)

\(=x^3+x^2+4x^2+4x+2x+2\)

\(=x^2\left(x+1\right)+4x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+4x+2\right)\)

28 tháng 10 2017

phần b là 6^4 nhé

27 tháng 10 2021

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

31 tháng 10 2021

1.\(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x+y\right)^2-\left(2z\right)^2\right]=5\left(x+y-2z\right)\left(x+y+2z\right)\)

2. \(=\left(-5x^2+15x\right)+\left(x-3\right)=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)

3. \(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)

4.\(=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)

5. \(=\left(x^2+x\right)+\left(3x+3\right)=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)

6. \(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)

7. \(=\left(x^2+x\right)-\left(5x+5\right)=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)

31 tháng 10 2021

\(1,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ 2,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ 3,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ 4,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=x^2+x+3x+3=\left(x+3\right)\left(x+1\right)\\ 6,=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ 7,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

30 tháng 4 2023

\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)

30 tháng 4 2023

\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)

2 tháng 5 2021

a, Để f(x) có nghiệm thì f(x) = 0

Hay: 4x2 - x = 0 ⇒ x(4x - 1) = 0 \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy...

b, Để f(x) có nghiệm thì f(x) = 0

Hay: x2 - 121 = 0 ⇒ x2 = 121 ⇒ \(\left[{}\begin{matrix}x=11\\x=-11\end{matrix}\right.\)

Vậy...

c, Để f(x) có nghiệm thì f(x) = 0

Hay: 5x + 2 = 0 \(\Rightarrow x=-\dfrac{2}{5}\)

Vậy...

d, Để đa thức có nghiệm thì 5x2 - 7x - 6 = 0

⇒ 5x2 - 10x + 3x - 6 = 0

⇒ 5x(x - 2) + 3(x - 2) = 0

⇒ (x - 2)(5x + 3) = 0

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{5}\end{matrix}\right.\)

Vậy...

a: x^2-7x+13=0

Δ=(-7)^2-4*1*13=49-52=-3<0

=>PTVN

b: -5x^2+5x-1.25=0

=>4x^2-4x+1=0

=>(2x-1)^2=0

=>2x-1=0

=>x=1/2

d: 2x^2+3x+1=0

=>(x+1)(2x+1)=0

=>x=-1 hoặc x=-1/2