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C5:

nC2H5OH = 8,2 / 46 = 0,2 (mol)

C2H5OH + 3O2-- (t^o)-- > 2CO2 + 3H2O

VCO2 = 0,4 . 22,4 = 8,96 (l)

VO2 = 0,6.22,4=13,44 (l)

\(\left(2^3\cdot9^4+9^3+45\right):\left(9^2\cdot10-9^2\right)\)

\(=\dfrac{9^3\cdot\left(2^3\cdot9+1\right)+45}{9^3}\)

\(=\dfrac{9^3\cdot73+45}{9^3}=\dfrac{5918}{81}\)

3: 

a: \(\Leftrightarrow x+1-6\sqrt{x+1}-9=0\)

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

=>x+1=9

=>x=8

b: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\left(\sqrt{\dfrac{1}{2}x+1}+3\right)}}=10\)

=>\(\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}-\dfrac{21}{4}}=10\)

=>\(\dfrac{1}{2}x-\dfrac{21}{4}-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}=100\)

=>\(\dfrac{7}{4}\cdot\sqrt{\dfrac{1}{2}x+1}=\dfrac{1}{2}x-\dfrac{21}{4}-100=\dfrac{1}{2}x-\dfrac{421}{4}\)

=>\(\sqrt{\dfrac{1}{2}x+1}=\dfrac{2}{7}x-\dfrac{421}{7}\)

=>1/2x+1=(2/7x-421/7)^2

=>1/2x+1=4/49x^2-1684/49x+177241/49

=>\(x\simeq249,77;x\simeq177,36\)

1B

2A

3C

4D

5A

15 C
Ex5:
1. B
2. A
3. C
4. D
5. A

15 C
C5:
1. B
2. A
3. C
4. D
5. A

B. Sew

Renew :làm mới 

a) 196:4-12.(-5)=49-(-60)=109