P = \(\dfrac{1}{a^2-ab+b^2}+\dfrac{3}{ab}=\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{ab}+\dfrac{1}{ab}+\dfrac{1}{ab}\)

\(\ge\dfrac{\left(1+1+1+1\right)^2}{a^2-ab+b^2+ab+ab+ab}=\dfrac{16}{\left(a+b\right)^2}=\dfrac{16}{2^2}=4\)

"=" khi a = b = 1 

(x + 2)(x - 3)(x² + 2x - 24) = 16x² 

<=> (x + 2)(x - 3)[(x + 1)² - 25] = 16x² 

<=> (x + 2)(x - 3)(x + 6)(x - 4) = 16x² 

<=> (x² + 8x + 12)(x² - 7x + 12) = 16x² 

<=> \(\left(x^2+0,5x+12+7,5x\right)\left(x^2+0,5x+12-7.5x\right)=16x^2\)

<=> \(\left(x^2+0,5x+12\right)^2-\left(7,5x\right)^2=16x^2\)

<=> \(\left(x^2+0,5x+12\right)^2=\left(8,5x\right)^2\)

<=> \(\left(x^2+9x+12\right)\left(x^2-8x+12\right)=0\)

<=> \(\left(x+\dfrac{9}{2}-\dfrac{\sqrt{33}}{2}\right)\left(x+\dfrac{9}{2}+\dfrac{\sqrt{33}}{2}\right)\left(x-2\right)\left(x-6\right)=0\)

<=>\(\left[{}\begin{matrix}x=\dfrac{\sqrt{33}-9}{2}\\x=\dfrac{-\sqrt{33}-9}{2}\\x=2\\x=6\end{matrix}\right.\)

Câu b) có (x;y) thỏa mãn : (x;y) = (0;19) 

1) whose 2) who ; 3) which ; 4) where 

5) that 6) which 7) whom ; 8) when  

\(x^2+1+3x=x\sqrt{x^2+1}+3\sqrt{x^2+1}\)

<=> \(\sqrt{x^2+1}\left(\sqrt{x^2+1}-x\right)-3\left(\sqrt{x^2+1}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\\\sqrt{x^2+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\\x^2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}∃x̸\\x=\pm\sqrt{8}\end{matrix}\right.\)

ĐKXĐ : \(x\ge2\)

Với \(A=\dfrac{x+3}{\sqrt{x}}\)

Khi đó \(A\sqrt{x}+x-1=2\sqrt{3x}+2\sqrt{x-2}\)

<=> \(\dfrac{x+3}{\sqrt{x}}.\sqrt{x}+x-1=2\sqrt{3x}+2\sqrt{x-2}\)

<=> \(x+1=\sqrt{3x}+\sqrt{x-2}\)

Đặt \(\sqrt{3x}=a;\sqrt{x-2}=b\left(a>0;b\ge0\right)\)

Khi đó \(a^2-b^2=2\left(x+1\right)\Leftrightarrow\dfrac{a^2-b^2}{2}=x+1\)

PT trở thành \(\dfrac{a^2-b^2}{2}=a+b\)

<=> \(\left(a+b\right)\left(\dfrac{a-b}{2}-1\right)=0\)

<=> \(\dfrac{a-b}{2}-1=0\left(a+b>0\right)\)

<=> a = b + 2 

Khi đó \(\sqrt{3x}=\sqrt{x-2}+2\)

<=> \(\left\{{}\begin{matrix}3x=x+2+4\sqrt{x-2}\\x\ge2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=2\sqrt{x-2}\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=4\left(x-2\right)\\x\ge2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x\ge2\end{matrix}\right.\Leftrightarrow x=3\)(tm) 

\(\)

1) In spite of the fact that I was late , I didn't miss the bus

2) Despite the fact that I was in a hurry , I didn't forget to lock the door 

3) In spite of heavy rain , they went to the football match .

4) In spite of necessary qualifications , she didn't get the job.

5) Despite the fact that she hates rock music,she went to the performance

6) In spite of the fact that they were very sucessful , they never made much money

7) Despite the fact that she knew the answer , she frefused to tell him 

8) Despite having a lot of money , she doesn't help the homeless

9) Despite the fact that I had planned very carefully, we ended up staying in a really bad hotel 

10) In spite of the fact that I wasn't keen on the film , I thought the music was beautiful 

d) Trích mẫu thử 5 ống nghiệm 

+) Cho nước vào 5 ống thu được 5 dd

+) Cho quỳ tím vào 5 ống trên 

Nếu không chuyển màu : CaCl₂ 

Nếu chuyển đỏ : P₂O₅ (P₂O₅ + H₂O ---> H₃PO4) 

Nếu chuyển xanh => Na₂O ; CaO ; Al₂O₃ 

(Na₂O + H₂O ---> NaOH ; CaO + H₂O ---> Ca(OH)₂ ; 

Al₂O₃ + H₂O ---> Al(OH)₃ 

Trong 3 base trên , base không tan => Al(OH)₃ 

base tan => Ca(OH)₂ ; NaOH

- Cho CaO ; Na₂O  tác dụng CO₂ 

CaO + CO₂ ---> CaCO₃ (không tan trong nước) 

Na₂O + CO₂ ---> Na₂CO₃ (tan trong nước) 

a) Trích mẫu thử vào 4 ống nghiệm

Cho quỳ tím vào 4 ống nghiệm 

+ Qùy tím đỏ => H₂SO₄ 

+ Qùy tím xanh => Ca(OH)₂

+) Không đổi màu : Na₂SO₃ ; H₂O

Cho H₂SO₄ tác dụng với nước và Na₂SO₃ 

+) Nếu pư có khí bay lên là Na₂SO₃

(H₂SO₄ + Na₂SO₃ ---> Na₂SO₄ + H₂O + SO₂ \(\uparrow\))

+) Không có khí bay lên là H₂O 

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{2xy}\)

Áp dụng BĐT Schwarz : \(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{\left(1+1\right)^2}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}=4\)

Lại có \(\dfrac{1}{2xy}=\dfrac{2}{4xy}\ge\dfrac{2}{\left(x+y\right)^2}=2\)

Cộng vế với vế được P \(\ge6\) ("=" khi x = y = 1/2)

Vậy Min P = 6 <=> x = y = 1/2