Bài 2: 

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

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

3) \(4x^2-9=\left(2x\right)^2-3^2=\left(2x+3\right)\left(2x-3\right)\)

4) \(9-25x^2=3^2-\left(5x\right)^2=\left(3-5x\right)\left(3+5x\right)\)

5) \(4x^2-25=\left(2x\right)^2-5^2=\left(2x+5\right)\left(2x-5\right)\)

6) \(9x^2-36=\left(3x\right)^2-6^2=\left(3x-6\right)\left(3x+6\right)\)

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

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

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

10) \(\left(3x\right)^2-9y^4=\left(3x\right)^2-\left(3y^2\right)^2=\left(3x-3y^2\right)\left(3x+3y^2\right)\)

Bài 2: 

21) \(\left(\dfrac{x}{3}-\dfrac{y}{4}\right)\left(\dfrac{x}{3}+\dfrac{y}{4}\right)=\left(\dfrac{x}{3}\right)^2-\left(\dfrac{y}{4}\right)^2=\dfrac{x^2}{9}-\dfrac{y^2}{16}\)

22) \(\left(\dfrac{x}{y}-\dfrac{2}{3}\right)\left(\dfrac{x}{y}+\dfrac{2}{3}\right)=\left(\dfrac{x}{y}\right)^2-\left(\dfrac{2}{3}\right)^2=\dfrac{x^2}{y^2}-\dfrac{4}{9}\)

23) \(\left(\dfrac{x}{2}+\dfrac{y}{3}\right)\left(\dfrac{x}{2}-\dfrac{y}{3}\right)=\left(\dfrac{x}{2}\right)^2-\left(\dfrac{y}{3}\right)^2=\dfrac{x^2}{4}-\dfrac{y^2}{9}\)

24) \(\left(2x-\dfrac{2}{3}\right)\left(\dfrac{2}{3}+2x\right)=\left(2x-\dfrac{2}{3}\right)\left(2x+\dfrac{2}{3}\right)=\left(2x\right)^2-\left(\dfrac{2}{3}\right)^2=4x^2-\dfrac{4}{9}\)

25) \(\left(2x+\dfrac{3}{5}\right)\left(\dfrac{3}{5}-2x\right)=\left(\dfrac{3}{5}+2x\right)\left(\dfrac{3}{5}-2x\right)=\left(\dfrac{3}{5}\right)^2-\left(2x\right)^2=\dfrac{9}{25}-4x^2\)

26) \(\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)\left(\dfrac{4}{3}+\dfrac{1}{2}x\right)=\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)\left(\dfrac{1}{2}x+\dfrac{4}{3}\right)=\left(\dfrac{1}{2}x\right)^2-\left(\dfrac{4}{3}\right)^2=\dfrac{1}{4}x^2-\dfrac{16}{9}\)

27) \(\left(\dfrac{2}{3}x^2-\dfrac{y}{2}\right)\left(\dfrac{2}{3}x^2+\dfrac{y}{2}\right)=\left(\dfrac{2}{3}x^2\right)^2-\left(\dfrac{y}{2}\right)^2=\dfrac{4}{9}x^4-\dfrac{y^2}{4}\)

28) \(\left(3x-y^2\right)\left(3x+y^2\right)=\left(3x\right)^2-\left(y^2\right)^2=9x^2-y^4\)

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

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

Bài `3`

\(1,\left(2x+1\right)^2+\left(2x-1\right)^2\\ \left[\left(2x\right)^2+4x+1^2\right]+\left[\left(2x\right)^2-4x+1^2\right]\\ =4x^2+4x+1+4x^2-4x+1\\ =8x^2+2\)

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

\(3,\left(x+2y\right)^2-\left(x-2y\right)^2\\ =\left[\left(x+2y\right)+\left(x-2y\right)\right]\left[\left(x+2y\right)-\left(x-2y\right)\right]\\ =\left(x+2y+x-2y\right)\left(x+2y-x+2y\right)\\ =2x.4y=8xy\)

\(4,\left(3x+y\right)^2+\left(x-y\right)^2\\ =\left[\left(3x\right)^2+6xy+y^2\right]+\left[\left(x^2-2xy+y^2\right)\right]\\ =6x^2+6xy+y^2+x^2-2xy+y^2\\ =7x^2+4xy+2y^2\)

\(5,-\left(x+5\right)^2-\left(x-3\right)^2\\ =-\left(x^2+10x+5^2\right)-\left(x^2-6x+3^2\right)\\ =-x^2-10x-25-x^2+6x-9\\ =-2x^2+16x-34\)

\(6,\left(3x-2\right)^2-\left(3x-1\right)^2\\ =\left[\left(3x-2\right)+\left(3x-1\right)\right]\left[\left(3x-2\right)-\left(3x-1\right)\right]\\ =\left(3x-2+3x-1\right)\left(3x-2-3x+1\right)\\ =\left(6x-3\right)\left(-1\right)=-6x+3\)

`@ Kidd`

bài này ko thể tính nhanh dc bạn ạ 

 kết quả bằng =3425839200

2037510

@tân nguyễn phuong vy sai cả hai

2039190

Bài 4.

a) 3xy² - 45x²y = 3xy( y - 15x )

b) 25y² - 4x² + 4x - 1

= 25y² - ( 4x² - 4x + 1 )

= ( 5y )² - ( 2x - 1 )²

= ( 5y - 2x + 1 )( 5y + 2x - 1 )

c) x² - 5x + xy - 5y

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

= ( x - 5 )( x + y )

d) x² - 8x - 33

= x² + 3x - 11x - 33

= x( x + 3 ) - 11( x + 3 )

= ( x + 3 )( x - 11 )

Bài 5.

a) A = ( x - 2 )³ - x²( x - 4 ) + 8

= x³ - 6x² + 12x - 8 - x³ + 4x² + 8

= -2x² + 12x

B = ( x² - 6x + 9 ) : ( x - 3 ) - x( x + 7 ) - 9

= ( x - 3 )² : ( x - 3 ) - x² - 7x - 9

= x - 3 - x² - 7x - 9

= -x² - 6x - 12

b) Với x = -1 thì A = -2.(-1)² + 12.(-1) = -2 - 12 = -14

Đề sai r bn

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x-3\\3x+2=3-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-5\\7x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{7}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2+3x=4x-3\\2+3x=3-4x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{7}\end{matrix}\right.\)

a) 142 + (-67) + 858 - 33

= 142 - 67 + 858 - 33

= (858 + 142) - (67 + 33)

= 1000 - 100

= 900

Gọi số lớn là a, số bé là b

(ĐIều kiện: \(b\ne0\))

Tổng của chúng là 59 nên a+b=59

Số lớn chia số bé được 6 dư 33 nên \(\dfrac{a}{b}=6\left(dư33\right)\)

=>a=6b+33

a+b=59

=>6b+33+b=59

=>7b=59-33=26

=>\(b=\dfrac{26}{7}\)

=>\(a=59-\dfrac{26}{7}=\dfrac{387}{7}\)

Vậy: Hai số cần tìm là \(\dfrac{387}{7};\dfrac{26}{7}\)

c: \(=\dfrac{4}{9}\left(\dfrac{5}{7}+\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{7}{5}\right)=\dfrac{4}{9}\cdot\left(1-1\right)=0\)

d: \(=\dfrac{4-3-1}{12}\cdot\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)=0\cdot\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)=0\)