a) 2x(x-3) + (3-x) = 0

2x(x - 3) - (x - 3) = 0

(2x - 1)(x - 3) = 0

=> 2x - 1 = 0 hoặc x - 3 = 0

\(x=\dfrac{1}{2}\) hoặc x = 3

b) 3x(x+5)-6(x+5)=0

(3x - 6)(x + 5) = 0

=> 3x - 6 = 0 hoặc x + 5 = 0

x = 2 hoặc x = -5

5(x² + 4x + 4) + 2(x² - 10x + 25) - 7(x² - 36)

= 5x² + 20x + 20 + 2x² - 20x + 50 - 7x² + 252

= 322

1) H₃PO₄ + KOH → K₃PO₄ + H₂O

2) NaOH + CO₂ → Na₂CO₃ + H₂O

10. will ask

11. will repair

Đề bài yêu cầu j?

Đề lỗi rồi kìa

1. B

2. C

3. B

4. D

5. documentary

6. voluntary

7. selection

8. unfairly

9. A

10. B

11. A

12. A

13. C

14. A

15. D

16. B

17. B

18. D

điều chỉnh chỗ x < 1 thành 0 < x < 1 nha

a) \(P=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(1-x\right)^2}{2}\)

\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(x-1\right)^2}{2}\)

\(=\dfrac{-2\sqrt{x}\left(x-1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)

\(=\dfrac{-2\sqrt{x}.\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)

\(=-\sqrt{x}.\left(\sqrt{x}-1\right)\)

b) Yêu cầu bài toán: \(-\sqrt{x}.\left(\sqrt{x}-1\right)>0\Leftrightarrow\sqrt{x}-1< 0\) (do \(-\sqrt{x}< 0\)) \(\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)

BĐT \(\Leftrightarrow\sqrt{2\left(a^2+b^2\right)}\ge a+b\)

\(\Leftrightarrow2a^2+2b^2\ge a^2+2ab+b^2\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)

Đẳng thức xảy ra \(\Leftrightarrow a=b\)

Ta có: đpcm