1. Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)=n_C\)

\(n_{H_2O}=\dfrac{8,1}{18}=0,45\left(mol\right)\) \(\Rightarrow n_H=2n_{H_2O}=0,9\left(mol\right)\)

⇒ m = m_C + m_H = 0,25.12 + 0,9.1 = 3,9 (g)

2. Ta có: n_H2O > n_CO2 ⇒ A là ankan.

→ CTPT chung của A là \(C_{\overline{n}}H_{2\overline{n}+2}\)

⇒ n_A = 0,45 - 0,25 = 0,2 (mol)

\(\Rightarrow M_A=\dfrac{3,9}{0,2}=19,5\left(g/mol\right)\)

\(\Rightarrow d_{A/H_2}=\dfrac{19,5}{2}=9,75\)

Có: \(12\overline{n}+2\overline{n}+2=19,5\Rightarrow\overline{n}=1,25\)

Mà: 2 chất có PTK hơn kém nhau 28 đvC → hơn kém 2 C.

⇒ 2 chất đó là CH₄ và C₃H₈.

6.C

7.A

8.B

đăng từng câu 1 thôi

Cái bpt nó dài hơn pt nên đăng 2 câu 1 thôi bạn

3)\(\dfrac{3x-1}{4}-\dfrac{3\left(x-2\right)}{8}-1>\dfrac{5-3x}{2}\)
<=> \(\dfrac{2\left(3x-1\right)}{8}-\dfrac{3\left(x-2\right)}{8}-\dfrac{8}{8}>\dfrac{4\left(5-3x\right)}{8}\)
=>\(6x-2-3x+6-8>20-12x\)
<=>\(6x-3x+12x>20+2-6+8\)
<=>15x>24
<=>x>\(\dfrac{8}{5}\)
vậy nghiệm của bất pt là x>8/5

\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)

\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)

7\(x\) < 36 < 63\(x\) + 7

⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)

\(\dfrac{29}{63}\)<  \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}

⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)

\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)

=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)

\(\Rightarrow7x< 36< 7x+7\)

\(\Rightarrow x< \dfrac{36}{7}< x+1\)

\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)

\(\Rightarrow x=5\)

tik cho mình nhé

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T

F(weak#strong)  

F(2#4)

F (march#january)

A  D

7 F (nation => nations)

8 F

9 F

10 F

B

11 A

12 D

6. Egypt is believed to be the driest country in the world.

7. The Taj is said to have been built with blind people who couldn't see how beautiful it is.

8. He is alleged to have kicked a policeman.

9. The train was supposed to arrive at 11.30.

10. The two injured men are thought to have been repairing overhead cables.

`7/8 - 3/5 =-x`

`=> x=-7/8 + 3/5`

`=> x= -35/40 + 23/40`

`=> x= -12/40`

`=> x= -3/10`

Vậy `x=-3/10`

a) Có \(\widehat{OAM}=90^0\) => Tam giác \(OAM\) nội tiếp đường tròn đường kính OM 

=> O,A,M cùng thuộc đường tròn đường kính OM (*)

Có \(\widehat{OBM}=90^0\) => Tam giác \(OBM\) nội tiếp đường tròn đường kính OM 

=> O,B,M cùng thuộc đường tròn đường kính OM (2*)

Do N là trung điểm của PQ => \(ON\perp PQ\)( Vì trong một đt, đường kính đi qua trung điểm của một dây ko đi qua tâm thì vuông góc với dây ấy)

=> \(\widehat{ONM}=90^0\) => Tam giác \(ONM\) nội tiếp đường tròn đường kính OM 

=> O,N,M cùng thuộc đt đường kính OM (3*)

Từ (*) (2*) (3*) => O,M,N,A,B cùng thuộc đt đk OM hay đt bán kính \(\dfrac{OM}{2}\)

b) Có AM//PS (cùng vuông góc với OA)

Gọi E là gđ của PS với (O) => \(sđ\stackrel\frown{AE}=sđ\stackrel\frown{AP}\)

Có \(\widehat{PRB}=\dfrac{1}{2}\left(sđ\stackrel\frown{AE}+sđ\stackrel\frown{PB}\right)\)\(=\dfrac{1}{2}\left(sđ\stackrel\frown{AP}+sđ\stackrel\frown{PB}\right)=\dfrac{1}{2}sđ\stackrel\frown{AB}\)

=> \(\widehat{PRB}=\widehat{MAB}=\dfrac{1}{2}sđ\stackrel\frown{AB}\)

Có BNAM nội tiếp => \(\widehat{MAB}=\widehat{MNB}\)

\(\Rightarrow\widehat{PRB}=\widehat{MNP}\) => PRNB nội tiếp

\(\Rightarrow\widehat{BRN}=\widehat{BPN}\) mà \(\widehat{BPN}=\widehat{BAQ}=\dfrac{1}{2}sđ\stackrel\frown{BQ}\)

\(\Rightarrow\widehat{BRN}=\widehat{BAQ}\) => RN//AQ hay RN // SQ mà N la trung điểm của PQ

=> RN là đường TB của tam giác PSQ

=> R là trung điểm của PS <=> PR=RS