\(n_{H_2}=0,2\left(mol\right)\)

\(2H_2+O_2\rightarrow2H_2O\)

0,2     0,1         0,2    (mol)

a) \(m_{H_2O}=0,2.18=3,6\left(g\right)\)

b) \(V_{O_2}=0,1.22,4=2,24\left(l\right)\)

A

4)a)\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{2}-\sqrt{3}\right|\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\left(do\sqrt{3}>\sqrt{2}\right)\)

\(=2\sqrt{2}\)

b)\(=\left|\sqrt{15}-4\right|-\left|3-\sqrt{15}\right|\)

\(=4-\sqrt{15}-\sqrt{15}+3\) (do \(4>\sqrt{15}>3\))

\(=7-2\sqrt{15}\)

3)a)\(=\dfrac{\sqrt{6}\left(\sqrt{6}+1\right)}{\sqrt{6}+1}-\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}\)

\(=\sqrt{6}-\sqrt{6}=0\)

b)\(=\dfrac{\sqrt{3}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{\sqrt{2}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}\)

\(=\sqrt{3}-\sqrt{2}\)

\(x+\dfrac{x}{2}=\dfrac{3}{4}\)

\(\dfrac{2x}{2}+\dfrac{x}{2}=\dfrac{3}{4}\)

\(\dfrac{3x}{2}=\dfrac{3}{4}\)

\(3x.4=2.3\)

\(12x=6\)

\(x=\dfrac{1}{2}\)

1C

2C

3A

4B

5D

6C

7B

8C

9A

10B

11C

12D

14A

15B

16D

17A

18C

19A

\(\left(2x-7y\right)^2\)

\(\left(6-x\right)^2\)

\(-\left(x+2y\right)^2\)

\(-\left(x-3\right)^2\)

\(\left(3-5x\right)^2\)

1)\(\left(x+2\right)^2\)

2)\(\left(x+3\right)^2\)

3)\(\left(2x+1\right)^2\)

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

5)\(\left(x-1\right)^2\)

6)\(\left(x-4\right)^2\)

7)\(\left(6-x\right)^2\)

8)\(\left(2x-3y\right)^2\)

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

a)\(S=\left(\dfrac{x+1+\sqrt{x}}{x+1}\right):\left(\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right)\) \(đk:x\ne\pm1\)

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

\(S=\dfrac{x+1+\sqrt{x}}{\sqrt{x}-1}\)

b)\(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\left(TMĐK\right)\)

\(\sqrt{x}=\sqrt{3}-1\)

Từ đó ta có :

\(S=\dfrac{4-2\sqrt{3}+1+\sqrt{3}-1}{\sqrt{3}-1-1}\)

\(S=-5-2\sqrt{3}\)