\(49-4x^2+2x+7\)

\(=-4x^2+2x+56\)

\(=-2\left(2x^2-x-28\right)\)

\(=-2\left(2x^2-8x+7x-28\right)\)

\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]\)

\(=-2\left(x-4\right)\left(2x+7\right)\)

\(49-4x^2+2x+7=-4x^2+2x+56\)

\(=-2\left(2x^2-x+28\right)=-2\left(2x^2-8x+7x+28\right)\)

\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]=-2\left(2x+7\right)\left(x-4\right)\)

= ( 2x - 5 )(2x + 5) - ( 2x-  5 )(2x + 7 )

= (2x - 5 ) [ 2x+  5 - ( 2x+  7 ) ] 

= ( 2x- 5 ) ( 2x + 5 - 2x - 7 )

= -  2( 2x- 5 ) 

\(4x^2-49=\left(2x\right)^2-7^2=\left(2x-7\right)\left(2x+7\right)\)

cho mình hỏi dấu ^ có ngĩa là gì

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

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

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

thầy/cô có thế giải 1 cách dễ hiểu hơn đc ko ạ

Bài 1: 

b: \(3x-6=x^2-16\)

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

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

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

\(4x^2-25+\left(2x+7\right).\left(5-2x\right)\)

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

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

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

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

#)Giải :

\(x^3-2x-4\)

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

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

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

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

\(x^4+2x^3+5x^2+4x-12\)

\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)

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

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

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

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

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

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

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

Bài 1:

a) \(7x^2\left(x^2-5x+1\right)=7x^4-35x^3+7x^2\)

b) \(\left(2x-3\right)\left(x+7\right)=2x^2+11x-21\)

Bài 2:

a) \(4x^2y-8x^3y^2=4x^2y\left(1-2xy\right)\)

b) \(2x-4y-ax+2ay=x\left(2-a\right)-2y\left(2-a\right)=\left(2-a\right)\left(x-2y\right)\)