x²-6x+5

=x²-5x-x+5

=x(x-5)-(x-5)

=(x-5)(x-1)

\(x^2-6x+5\)

\(=x^2-x-5x+5\)

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

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

Ta có: \(8-27x^6y^3\)

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

\(=\left(2-3x^2\right)\left(4+6x^2y+9x^4y^2\right)\)

\(8-27x^6y^3\)

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

\(=\left(2-3x^2y\right)\left(4+6x^2y+9x^2y^2\right)\)

Thank mấy bn :)

b) \(\dfrac{2}{3}x^2y-2xy^2+4xy=2xy\left(\dfrac{1}{3}x-y+2\right)\)

c) \(4x^2-2x-3y-9y^2\)

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

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

a) Áp dụng hằng đằng thức hiệu của 2 bình phương ta có

\(x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

                                                       Bài giải

\(a,\text{ }x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

\(b,\text{ }4x-5=\left(2x\right)^2-\left(\sqrt{5}\right)^2=\left(2x-\sqrt{5}\right)\left(2x+\sqrt{5}\right)\)

\(x^2-7=x^2-\sqrt{7}^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

\(4x-5=\left(2\sqrt{x}\right)^2-\sqrt{5}^2=\left(2\sqrt{x}-\sqrt{5}\right)\left(2\sqrt{x}+\sqrt{5}\right)\)

Answer:

\(81x^4+4\)

\(=81x^4+36x^2+4-36x^2\)

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

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

(1+x2)2−4x(1−x2)

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

đặt \(\left(1-x^2\right)\)= a

ta có :

- a . a - 4x .a

= a ( - a - 4x )

thay a = \(\left(1+x^2\right)\) ta có

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

phân tích tiếp nhé !
 

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