A) \(\left(x-3\right)^2-\left(x+2\right)^2\)

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

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

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

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

\(=8x^3-y^3-8x^3-y^3\)

\(=-2y^3\)

C) \(x^2+6x+8\)

\(=x^2+6x+9-1\)

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

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

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

bài 3 A) \(x^2-16=0\)

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

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

vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

B) \(x^4-2x^3+10x^2-20x=0\)

\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\left(x^3+10x\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

x=0

x=2

a: =-3x^2y*x^2y+3x^2y*2xy

=-3x^4y^2+6x^3y^2

b: =x^3-x^2y+x^2y+y^2=x^3+y^2

c: =x*4x^3-x*5xy+2x*x

=4x^4-5x^2y+2x^2

d: =x^3+x^2y+2x^3+2xy

=3x^3+x^2y+2xy

\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)

\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)

\(=-5x-27\)

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

\(=8x^3+y^3-\left(8x^3-y^3\right)\)

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

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

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

\(=z^2\)

a)

=\(x^3-6x^2+12x+8-27-x^3+x+6x^2-18x\) 

=-5x-19

b)

=\(8x^3+y^3-8x^3+y^3\) 

=\(2y^3\) 

c)

=(x+y+z-x-y)\(^2\) +x+y

=\(z^2+x+y\) 

hc tốt

1, \(2x^2+4x=2x\left(x+2\right)\)

2, \(15x^3+5x^2-10x=5x\left(3x^2+x-2\right)=5x\left(x-\dfrac{2}{3}\right)\left(x+1\right)\)

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

4) \(3\left(x-y\right)+5x\left(y-x\right)=\left(x-y\right)\left(3-5x\right)\)

5) \(5x^2-10x=5x\left(x-2\right)\)

6) \(3x-6y=3\left(x-2y\right)\)

7) \(25x^2+5x^3+x^2y=x^2\left(25+5x+y\right)\)

8) \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)

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

10) \(10x\left(x-y\right)-8y\left(y-x\right)=\left(10x+8y\right)\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)

\(1,=2x\left(x+2\right)\\ 2,=5x\left(3x^2+x-2\right)\\ 3,=\left(x-2y\right)\left(5x^2+15x\right)=5x\left(x+3\right)\left(x-2y\right)\\ 4,=\left(x-y\right)\left(3-5x\right)\\ 5,=5x\left(x-2\right)\\ 6,=3\left(x-2y\right)\\ 7,=5x^2\left(5+x+y\right)\\ 8,=7xy\left(2x-3y+4xy\right)\\ 9,=\left(y-1\right)\left(x-y\right)\\ 10,=\left(x-y\right)\left(10x+8y\right)=2\left(5x+4y\right)\left(x-y\right)\)

a, (3 - \(x\))(4y + 1) = 20

   Ư(20) = { -20; -10; -5; -4; -2; -1; 1; 2; 4; 5; 10; 20}

Lập bảng ta có:

Vậy các cặp \(x;y\) nguyên thỏa mãn đề bài là:

(\(x;y\)) =(-1; 1); (-17; 0)

b, \(x\left(y+2\right)\)+ 2\(y\) = 6

    \(x\) = \(\dfrac{6-2y}{y+2}\)

\(x\in\) Z ⇔ 6 - \(2y⋮\) \(y\) + 2 ⇒-(2y + 4) +10 ⋮ \(y\) + 2 ⇒ -2(\(y\)+2) +10 ⋮ \(y\)+2

⇒ 10 ⋮ \(y\) + 2

Ư(10) = { -10; -5; -2; -1; 1; 2; 5; 10}

Lập bảng ta có:

 Theo bảng trên ta có các cặp \(x;y\)

 nguyên thỏa mãn đề bài lần lượt là:

(\(x;y\)    ) =(-3; -12); (-4; -7); (-12; -3); (8; -1); (3; 0); (0;3 (-1; 8)                           

câu 1:

a,x²+2x-4z²+1

=x²+2x.1+1²-(2z)²

=(x+1)²-(2z)²

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

bạn nên dùng hằng đẳng thức đã học

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

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

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

b) \(4x^2+4xy-16+y^2=\left(4x^2+4xy+y^2\right)-16\)

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

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

c)

a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

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

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

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

c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

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

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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

a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)

Thay x = 15 vào bt A ta có

A = 9 . 15 = 135

b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)

Thay x = -1/5 ; y = - 1/2 vào bt B ta có

\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(=9x^2y^2-xy^3-8x^3\)

Thay x = 1/2 ; y = 2 vào bt C ta có

\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)

d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)

\(=12x^2+12x-3\)

\(\left|x\right|=2\Rightarrow x=\pm2\)

Thay x = 2 vào bt D có

\(D=12.4+12.2-3=69\)

Thay x = - 2 vào bt D ta có

\(D=12.4-12.2-3=21\)