`{(-6x + 3y = -3),(-10x - y = -5 - 3xy):}`

`<=> {(-2x + y = -1),(-10x - y = -5 - 3xy):}`

`<=> {(y =2x -1),(-10x - y = -5 - 3xy):}`

`<=> {(y =2x -1),(-10x - (2x -1) = -5 - 3x(2x -1)(1)):}`

Từ (1) `<=> -10x - 2x + 1 = -5x - 6x^2 + 3x`

`<=> 6x^2 - 3x + 5 -10x - 2x + 1 = 0 `

`<=> 6x^2 - 15x + 6 = 0`

`<=> 2x^2 - 5x + 2 = 0`

`<=> (2x^2 - 4x) - (x - 2) = 0`

`<=> 2x(x-2) - (x-2) = 0`

`<=> (2x - 1)(x-2) = 0`

<=> \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

Hệ phương trình <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\\y=2x-1\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

11) expands

12) will be cold

14) would you do

15) water

16) give

17) wouldn't have taken 

18) won't find

20) notify

11 expands

12 will be

13 would have been

14 would you do

15 water

16 gave

17 wouldn't have take

18 won't find

19 do

20 notify

c: \(P=\frac{\sqrt{x^3}-\sqrt{y^3}}{x-y}-\frac{x}{\sqrt{x}+\sqrt{y}}-\frac{y}{\sqrt{y}-\sqrt{x}}\)

\(=\frac{x\cdot\sqrt{x}-y\sqrt{y}}{x-y}-\frac{x}{\sqrt{x}+\sqrt{y}}+\frac{y}{\sqrt{x}-\sqrt{y}}\)

\(=\frac{\left(x\sqrt{x}-y\sqrt{y}\right)-x\left(\sqrt{x}-\sqrt{y}\right)+y\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\frac{x\sqrt{x}-y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}+y\sqrt{y}}{x-y}=\frac{x\sqrt{y}+y\cdot\sqrt{x}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{NaOH}=0,3.1=0,3\left(mol\right)\)

\(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,3}{0,2}=1,5< 2\)

Pứ tạo muối axit trước.

\(CO_2+NaOH\rightarrow NaHCO_3\)

0,2----> 0,2 ----->  0,2

\(NaOH+NaHCO_3\rightarrow Na_2CO_3+H_2O\)

0,1------> 0,1 ----------> 0,1

Giả sử thể tích dung dịch thay đổi không đáng kể.

\(CM_{NaHCO_3}=\dfrac{0,2-0,1}{0,3}=\dfrac{1}{3}\left(M\right)\)

\(CM_{Na_2CO_3}=\dfrac{0,1}{0,3}=\dfrac{1}{3}\left(M\right)\)

1. He would have solved the puzzle had he watched the news.
2. If I had had a mobile, I could have phoned you.
3. Were I to ask you to lend me your dictionary, would you do it?
4. The money will only be paid should a new contract be signed.
5. I’d appreciate it if you would reply at your earliest convenience.
6. If only I had known you earlier.
7. But for your absent-mindedness then, the soup would have tasted excellent.
8. They would have paid less if they had booked the tickets yesterday.
9.If you like, you can stay for two days.
10.If the parents were to buy the cat, their children would be very happy.

As long as you follow the instructions carefully, there won't be any problems.

0 when

13 she

14 given

15 was

16 say

17 reached

18 who

19 somewhere

20 possible

21 girl

22 into

23 to 

24 was

ĐKXĐ: \(0\le x\le5\)

\(A=\dfrac{1}{\sqrt{35}}.\left(\sqrt{5x\left(35-7x\right)}+\sqrt{7x\left(35-5x\right)}\right)\)

\(A\le\dfrac{1}{2\sqrt{35}}\left(5x+35-7x+7x+35-5x\right)\)

\(A\le\sqrt{35}\)

\(A_{max}=\sqrt{35}\) khi \(\left\{{}\begin{matrix}5x=35-7x\\7x=35-5x\end{matrix}\right.\) \(\Rightarrow x=\dfrac{35}{12}\)