Trả lời:

7, 49y² - x² + 6x - 9

= 49y² - ( x² - 6x + 9 )

 = ( 7y )² - ( x - 3 )²

= ( 7y - x + 3 ) ( 7y - x - 3 )

8, sửa đề: 25x² - 4y² - 4y - 1

= 25x² - ( 4y² + 4y + 1 )

= ( 5x )² - ( 2y + 1 )

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

9, 4x² - y² + 8y - 16

= 4x² - ( y² - 8y + 16 )

= ( 2x )² - ( y - 4 )²

= ( 2x - y + 4 ) ( 2x + y - 4 )

a, \(49y^2-x^2+6x-9=49y^2-\left(x-3\right)^2=\left(7y-x+3\right)\left(7y+x-3\right)\)

b, đề sai rồi bạn 

c, \(4x^2-y^2+8y-16=4x^2-\left(y-4\right)^2=\left(2x-y+4\right)\left(2x+y-4\right)\)

(2x +y-4)

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

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

b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)

c, \(5x-2-25x^2+10x=0\)

\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)

= 1/5 nha

a) \(\left(6x-5y\right)^2=36x^2-60xy+25y^2\)

b) \(\left(4x-1\right)^2=16x^2-8x+1\)

c) \(\left(x+2\right)^2=x^2+4x+4\)

d) \(x^2-64=\left(x-8\right)\left(x+8\right)\)

e) \(4x^2-64=\left(2x-8\right)\left(2x+8\right)\)

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

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

h) \(\left(x-3\right)^3=x^3-9x^2+27x-27\)

k) \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

l) \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)

y) \(27y^3-1=\left(3y-1\right)\left(9y^2+3y+1\right)\)

^{ai Giúp mình với}

ko có đề bài hả bn

1, \(3x\left(x-7\right)+2x-14=0\)

\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)

2, \(x^3+3x^2-\left(x+3\right)=0\)

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

\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)

3, \(15x-5+6x^2-2x=0\)

\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)

\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)

4, \(5x-2-25x^2+10x=0\)

\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)

\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)

\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)

\(1,a^2-2a+1-b^2\)

\(=\left(a^2-2a+1\right)-b^2\)

\(=\left(a-1\right)^2-b^2\)

\(=\left(a-1-b\right)\left(a-1+b\right)\)       Khai triển thành hằng đẳng thức số 3 e  nhé.

\(2,x^2+2xy+y^2-81\)

\(=\left(x^2+2xy+y^2\right)-81\)

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

\(=\left(x+y-9\right)\left(x+y+9\right)\)Cái này cũng HĐT số 3 nè

\(3,x^2+6y-9-y^2\)

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

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

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

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

\(5,4x^2+y^2-9-4xy\)

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

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

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

Học tốt

ghytyyj

a: \(\Leftrightarrow A=25x^2y-13xy^2+y^3-11x^2y+2y^3=14x^2y-13xy^2+3y^3\)

b: \(A=-12x^4+15x^2y-2xy^2-7\)

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

2) \(9y^2-6y+1-25x^2=\left(3y\right)^2-2.3y+1-\left(5x\right)^2=\left(3y-1\right)^2-\left(5x\right)^2\)

\(=\left(3y-1-5x\right)\left(3y-1+5x\right)\)

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