Question

viết biểu thức sau dới dạng tích 

a/ (a^2-b^2)^2-(a^2+b^2)^2

b/ a^6-b^6

c/ -4x^3+9y^2

d/ (x+1)^3-(2-x)^3

e/ 8+(4x-3)^3

g/ 81-(9-x^2)^2

Answer 1

a. (a² - b²)² - (a² + b²)²

= (a² - b² - a² - b²)(a² - b² + a² + b²)

= -2b² . 2a²

b. a⁶ - b⁶

<=> (a³)² - (b³)²

<=> (a³ - b³)(a³ + b³)

Answer 2

\(a,\left(a^2-b^2\right)^2-\left(a^2+b^2\right)^2\\ =a^4-2a^2b^2+b^4-a^4-2a^2b^2-b^4\\ =-4a^2b^2\)

\(b,a^6-b^6=a^2\left(a^3-b^3\right)=a^2\left(a-b\right)\left(a^2+ab+b^2\right)\)

\(c,-4x^2+9y^2=\left(3y-2x\right)\left(3y+2x\right)\\ d,\left(x+1\right)^3-\left(2-x\right)^3\\ =\left(x+1-2+x\right)\left[\left(x+1\right)^2+\left(x+1\right)\left(2-x\right)+\left(2-x\right)^2\right]\\ =\left(2x-1\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)\\ =\left(2x-1\right)\left(x^2-x+7\right)\)

\(e,8+\left(4x-3\right)^3\\ =\left(8+4x-3\right)\left[64-8\left(4x-3\right)+\left(4x-3\right)^2\right]\\ =\left(4x+5\right)\left(64-32x+24+16x^2-24x+9\right)\\ =\left(4x+5\right)\left(16x^2-56x+97\right)\)

\(g,81-\left(9-x^2\right)^2\\ =\left(9-9+x^2\right)\left(9+9-x^2\right)\\ =x^2\left(18-x^2\right)\left[=x^2\left(\sqrt{18}-x\right)\left(\sqrt{18}+x\right)\right]\)

Chỗ trong ngoặc nếu bạn chưa học căn thì ko cần ghi nha

Answer 3

a: Ta có: \(\left(a^2-b^2\right)^2-\left(a^2+b^2\right)^2\)

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

\(=-4a^2b^2\)

b: \(a^6-b^6\)

\(=\left(a^3-b^3\right)\left(a^3+b^3\right)\)

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

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

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

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

\(=\left(2x-1\right)\left(x^2-x+7\right)\)