Bài 3

a) 2x(x - 3) - x + 3 = 0

2x(x - 3) - (x - 3) = 0

(x - 3)(2x - 1) = 0

x - 3 = 0 hoặc 2x - 1 = 0

*) x - 3 = 0

x = 3

*) 2x - 1 = 0

2x = 1

x = 1/2

Vậy x = 1/2; x = 3

b) (3x - 1)(2x + 1) - (x + 1)² = 5x²

6x² + 3x - 2x - 1 - x² - 2x - 1 - 5x² = 0

(6x² - x² - 5x²) + (3x - 2x - 2x) = 0 + 1 + 1

-x = 2

x = -2

Bài 2

a) 5x² + 30y

= 5(x² + 6y)

b) x³ - 2x² - 4xy² + x

= x(x² - 2x - 4y² + 1)

= x[(x² - 2x + 1) - 4y²]

= x[(x - 1)² - (2y)²]

= x(x - 1 - 2y)(x - 1 + 2y)

_{Chiều dài của hình chữ nhật sau khi tăng thêm 5% là:}

     _{100%+5%=105% (chiều dài ban đầu)}

_{Chiều rộng của hình chữ nhật sau khi tăng thêm 8% là:}

     _{100%+8%=108% (chiều rộng ban đầu)}

_{Diện tích của hình chữ nhật sau khi tăng là:}

     _{105%×108%=113,4% (diện tích ban đầu)}

              _{Đáp số:113,4$}

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

bài 11

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

b)

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

c)

\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

bài 12

\(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x=26\\ x=-2\)

b)

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Bài 1:

a) 2x^2 -3x + 1 = 2x^2 -2x -x +1 = 2x.(x-1) - (x-1) = (x-1).(2x-1)

b) 2x^3y - 2xy^3 - 4xy^2 - 2xy = 2xy.(x^2 - y^2 - 2y -1) = 2xy.[ x^2 - (y^2 + 2y+1)] = 2xy.[x^2 - (y+1)^2]

= 2xy.(x-y-1).(x+y+1)

Bài 1:

a) \(2x^2-3x+1\)

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

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

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

b) \(2x^2y-2xy^3-4xy^2-2xy\)

\(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)

c) Đặt \(C=\left(x^2+x+3\right)\left(x^2+x+5\right)-8\)

Đặt \(a=x^2+x+4\)

\(\Leftrightarrow C=\left(a-1\right)\left(a+1\right)-8\)

\(\Leftrightarrow C=a^2-1^2-8\)

\(\Leftrightarrow C=a^2-9\)

\(\Leftrightarrow C=\left(a-3\right)\left(a+3\right)\)

Thay \(a=x^2+x+4\)vào C ta có :

\(C=\left(x^2+x+4-3\right)\left(x^2+x+4+3\right)\)

\(C=\left(x^2+x+1\right)\left(x^2+x+7\right)\)

Bài 2:

a) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+2x\right)=0\)

\(x^3+8-x^3-2x=0\)

\(8=2x\)

\(x=4\)

b) \(x^2-2x-8=0\)

\(x^2-4x+2x-8=0\)

\(x\left(x-4\right)+2\left(x-4\right)=0\)

\(\left(x-4\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)

c) (x^2 + x+3).(x^2 + x +5) - 8 = (x^2+x+4-1).(x^2+x+4+1) - 8 = (x^2+x+4)^2 - 1 - 8 = (x^2+x+4)^2 - 3^2

= (x^2+x+4-3).(x^2+x+4+3) = (x^2+x+1).(x^2+x+7)

Bài 2:

a) (x+2).(x^2-2x+4) - (x^3+2x) = 0

x^3 + 8 - x^3 - 2x = 0

8 - 2x = 0

x = 4

b) x^2 - 2x - 8 = 0

x^2 +2x - 4x - 8 = 0

x.(x+2) - 4.(x+2) = 0

(x+2).(x-4) = 0

...

bn tự làm tiếp nha

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

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

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

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)

ỳtct7ct7c7c7t79tc9

1.a) 2x⁴-4x³+2x²

=2x²(x²-2x+1)

=2x²(x-1)²

b) 2x²-2xy+5x-5y

=2x(x-y)+5(x-y)

=(2x+5)(x-y)

2.

a) 4x(x-3)-x+3=0

=>4x(x-3)-(x-3)=0

=>(4x-1)(x-3)=0

=> 2 TH:

*4x-1=0            *x-3=0

=>4x=0+1        =>x=0+3

=>4x=1           =>x=3

=>x=1/4

vậy x=1/4 hoặc x=3

b) (2x-3)^2-(x+1)^2=0

=> (2x-3-x-1).(2x-3+x+1)=0

=>(x-4).(3x-2)=0

=> 2 TH

*x-4=0

=> x=0+4

=> x=4

*3x-2=0

=>3x=0-2

=>3x=-2

=>x=-2/3 

vậy x=4 hoặc x=-2/3

sửa 1 chút phần cuối:

3x-2=0

=>3x=0+2

=>3x=2

=>x=2/3

vậy x=2/3 hoặc....

a: =64x^4+16x^2y^2+y^4-16x^2y^2

=(8x^2+y^2)^2-(4xy)^2

=(8x^2+y^2-4xy)(8x^2+y^2+4xy)

b: =x^8+2x^4+1-x^4

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

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

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

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

c: =(x+1)(x^2-x+1)+2x(x+1)

=(x+1)(x^2-x+1+2x)

=(x+1)(x^2+x+1)

d: =(x^2-1)(x^2+1)-2x(x^2-1)

=(x^2-1)(x^2-2x+1)

=(x-1)^2*(x-1)(x+1)

=(x+1)(x-1)^3

1,

a, = 2x.(x-2)

b, = (x^2+y^2+2xy)-(2x+2y)

= (x+y)^2-2.(x+y)

= (x+y).(x+y-2)

2,

a,<=> x^2-1-x^2-2x = 3

 <=> -2x-1=3

<=> -2x=4

<=> x=4 : (-2) = -2

b, <=>(x^2-4x+4)-7=0

<=>(x-2)^2-7=0

<=> (x-2)^2=7

=> x-2=+-\(\sqrt{7}\)

<=> x=2+-\(\sqrt{7}\)

k mk nha

a, \(2x-4x\)

\(=-2x\)

b, \(x^2+y^2+2xy-2x-2y\)

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

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

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

\(\Leftrightarrow x^2-1-x^2-2x=3\)

\(\Leftrightarrow-2x=4\)

\(\Leftrightarrow x=-2\)

b,\(x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

1)

a)\(2x-4x=x\left(2-4\right)=-2x\)

b)\(x^2+y^2+2xy-2x-2y\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)\left(x+y-2\right)\)

2)

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

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

\(\Leftrightarrow\left(x^2-1\right)-x^2-2x=3\)

\(\Leftrightarrow x^2-1-x^2-2x=3\)

\(\Leftrightarrow-1-2x=3\)

\(\Leftrightarrow-2x=4\)

\(\Leftrightarrow=-2\)

b)\(x^2-4x+3=0\)

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

P/ s tham khảo nha