Question

Giúp em với ạ

Answer 1

Bài 1 bạn phân tích ra HĐT số 3 nhé

Answer 2

Bài 1: 

1) \(9x-4=\left(3\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)\)

2) \(25x-4=\left(5\sqrt{x}-2\right)\left(5\sqrt{x}+2\right)\)

3) \(1-16x=\left(1-4\sqrt{x}\right)\left(1+4\sqrt{x}\right)\)

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

5) \(9x-16=\left(3\sqrt{x}-4\right)\left(3\sqrt{x}+4\right)\)

6) \(x-25=\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)\)

Answer 3

Bài 3:

d) Ta có: \(\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)

\(=\left(x^2+x+4\right)^2+3x\left(x^2+x+4\right)+5x\left(x^2+x+4\right)+15x^2\)

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

\(=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)

\(=\left(x+2\right)^2\cdot\left(x^2+6x+4\right)\)

e) Ta có: \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)-24\)

\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)