Question

Bài 1: Viết các biểu thức dưới dạng tích ( cho x ≥ 0)

1) x² - 9 ; 2) 9x² - 16 ; 3) x²-5 ; 4) x - 9; 5) x- 3; 6) x + \(2\sqrt{x}+1\); 7) \(x-4\sqrt{x}+4\); 8) \(4x+4\sqrt{x}+1\); 9) \(x+2\sqrt{x}-35\)

Mọi người giúp em mk với ak, cảm ơn.....!!!!

Answer 1

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

2) \(9x^2-16=\left(3x\right)^2-4^2=\left(3x-4\right)\left(3x+4\right)\)

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

4) \(x-9=\left(\sqrt{x}\right)^2-3^2=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)(ĐK: \(x\ge0\))

5) \(x-3=\left(\sqrt{x}\right)^2-\left(\sqrt{3}\right)^2=\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)\)(ĐK: nt)

6) \(x+2\sqrt{x}+1=\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot1+1=\left(\sqrt{x}+1\right)^2\)(ĐK: nt)

7) \(x-4\sqrt{x}+4=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot2+2^2=\left(\sqrt{x}-2\right)^2\)(ĐK: nt)

8) \(4x+4\sqrt{x}+1=\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot1+1=\left(2\sqrt{x}+1\right)^2\)(ĐK:nt

9)

\(x+2\sqrt{x}-35\\ =x-5\sqrt{x}+7\sqrt{x}-35\\ =\sqrt{x}\left(\sqrt{x}-5\right)+7\left(\sqrt{x}-5\right)\\=\left(\sqrt{x}-5\right)\left(\sqrt{x}+7\right)\)(ĐK: nt)