a: Xét ΔCDE vuông tại D và ΔCAB vuông tại A có

\(\hat{DCE}\) chung

Do đó: ΔCDE~ΔCAB

=>\(\frac{CD}{CA}=\frac{CE}{CB}\)

=>\(CD\cdot CB=CE\cdot CA\)

b: AD là phân giác của góc BAC

=>\(\hat{BAD}=\hat{CAD}=\frac12\cdot\hat{BAC}=45^0\)

\(\frac{CD}{CA}=\frac{CE}{CB}\)

=>\(\frac{CD}{CE}=\frac{CA}{CB}\)

Xét ΔCDA và ΔCEB có

\(\frac{CD}{CA}=\frac{CA}{CB}\)

góc DCA chung

Do đó: ΔCDA~ΔCEB

=>\(\hat{CDA}=\hat{CEB}\) ; \(\hat{CAD}=\hat{CBE}\)

Gọi I là giao điểm của AD và BE

Xét ΔIAE và ΔIBD có

\(\hat{IAE}=\hat{IBD}\)

\(\hat{AIE}=\hat{BID}\) (hai góc đối đỉnh)

Do đó: ΔIAE~ΔIBD

=>\(\frac{IA}{IB}=\frac{IE}{ID}\)

=>\(\frac{IA}{IE}=\frac{IB}{ID}\)

Xét ΔIAB và ΔIED có

\(\frac{IA}{IE}=\frac{IB}{ID}\)

\(\hat{AIB}=\hat{EID}\) (hai góc đối đỉnh)

Do đó: ΔIAB~ΔIED

=>\(\hat{IAB}=\hat{IED}\)

=>\(\hat{IED}=45^0\)

Xét ΔDEB vuông tại D có \(\hat{DEB}=45^0\)

nên ΔDEB vuông cân tại D

=>DE=DB

c: Xét ΔBDF vuông tại D và ΔBAC vuông tại A có

\(\hat{DBF}\) chung

Do đó: ΔBDF~ΔBAC

=>\(\frac{BD}{BA}=\frac{BF}{BC}\)

=>\(\frac{BD}{BF}=\frac{BA}{BC}\)

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=12^2+16^2=144+256=400=20^2\)

=>BC=20(cm)

Xét ΔBDA và ΔBFC có

\(\frac{BD}{BF}=\frac{BA}{BC}\)

góc DBA chung

Do đó: ΔBDA~ΔBFC

=>\(\frac{AD}{FC}=\frac{BA}{BC}=\frac{12}{20}=\frac35\)

a) \(n_{NaOH}=C_{M,NaOH}\cdot V_{NaOH}=2\cdot0,2=0,4\left(mol\right)\)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\) 

Theo PTHH: \(n_{HCl}=n_{NaOH}=0,4\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{n_{HCl}}{C_{M,HCl}}=\dfrac{0,4}{1,2}=\dfrac{1}{3}\left(l\right)\)  

b) Thể tích dung dịch sau phản ứng là:

\(0,2+\dfrac{1}{3}=\dfrac{8}{15}\left(l\right)\)

Theo PTHH: \(n_{NaCl}=n_{NaOH}=0,4\left(mol\right)\) 

\(\Rightarrow C_{M,NaOH}=\dfrac{0,4}{\dfrac{8}{15}}=0,75\left(M\right)\) 

dòng cuối là NaCl :)))

a: Xét ΔAME và ΔADC có

\(\hat{AME}=\hat{ADC}\) (hai góc đồng vị, ME//DC)

\(\hat{MAE}\) chung

Do đó: ΔAME~ΔADC

=>Tỉ số đồng dạng là \(\frac{AE}{AC}=\frac13\)

Ta có: AE+EC=AC

=>\(EC=AC-AE=AC-\frac13AC=\frac23AC\)

Xét ΔCNE và ΔADC có

\(\hat{NCE}=\hat{DAC}\) (hai góc so le trong, NC//AD)

\(\hat{ACD}=\hat{CEN}\) (hai góc so le trong, EN//DC)

Do đó: ΔCNE~ΔADC

=>Tỉ số đồng dạng là \(\frac{CE}{AC}=\frac23\)

Xét ΔBAC và ΔDCA có

\(\hat{BAC}=\hat{DCA}\) (hai góc so le trong, AB//CD)

\(\hat{ABC}=\hat{CDA}\) (ABCD là hình bình hành)

Do đó: ΔBAC~ΔDCA

=>Tỉ số đồng dạng là \(\frac{BA}{DC}=1\)

b: E là trung điểm của MN

=>EM=EN

=>\(\frac{EM}{DC}=\frac{EN}{DC}\)

mà DC=AB

nên \(\frac{EM}{DC}=\frac{EN}{AB}\) (2)

Xét ΔADC có ME//DC
nên \(\frac{ME}{DC}=\frac{AE}{AC}\) (1)

Xét ΔCAB có NE//AB

nên \(\frac{NE}{AB}=\frac{CE}{CA}\) (3)

Từ (1),(2),(3) suy ra AE=CE

=>E là trung điểm của AC

\(n_{P_2O_5}=\dfrac{22,72}{142}=0,16mol\\ 4P+5O_2\xrightarrow[]{t^0}2P_2O_5\\ a.n_P=0,16\cdot\dfrac{4}{2}=0,32mol\\ m=m_P=0,32.31=9,92g\\ b.n_{O_2}=0,16\cdot\dfrac{5}{2}=0,4mol\\ 0V=V_{O_2}=0,4.22,4=8,96l\)

- A: Do you find that it's hotter these days?
- B: Yes, it's hot because it's the beginning of summer.
- A: No, I mean this year is hotter than last year.
- B: Maybe you're right because the earth's temperature increases year by year. This won't stop until people can reduce all types of pollution which causes the global warming.
- A: I think reducing pollution is not easy at all. Human life requires the producing of goods, so natural resources are exploited to make products. Then waste will still be created and people will pollute the environment again and again.
- B: Well, it's not easy, but it's possible. Many countries are trying to recycle things and look for new sources of energy. They also issue laws and have lots of programs to protect the environment.
- A: How can recycling things and finding new sources of energy help to reduce pollution?
- B: If people recycle more things, they will use fewer natural resources and there will be less waste. New sources of energy can replace the using of fossil fuels which results in most types of pollution.
- A: That seems the logical things. I hope it will be better soon because we can't live healthily in this hot weather and the polluted environment. I can't imagine how life on earth will look like in the next 100 years.
- B: Yeah, it's hard to imagine.

Mua khoá học thì có nha bạn...

W-ai? Li-ve a co-ment

đề khó hiểu quá

Chỉ cần hỏi một câu thôi là biết có chỗ trống để ngồi hay không?

Xin lỗi, mọi người làm ơn cho hỏi có còn chỗ trống để ngồi đợi xe bus không a?

Before:

Build an emergency kit and make a family communications plan.

Avoid building in a floodplain unless you elevate and reinforce your home.

Elevate the furnace, water heater and electric panel in your home if you live in an area that has a high flood risk.

Consider installing "check valves" to prevent flood water from backing up into the drains of your home.

If feasible, construct barriers to stop floodwater from entering the building and seal walls in basements with waterproofing compounds.

During:

Listen to the radio or television for information.

Be aware that flash flooding can occur. If there is any possibility of a flash flood, move immediately to higher ground. Do not wait for instructions to move.

Be aware of stream, drainage channels, canyons and other areas known to flood suddenly. Flash floods can occur in these areas with or without typical warnings such as rain clouds or heavy rain.

After:

Avoid moving water.

Stay away from damaged areas unless your assistance has been specifically requested by police, fire, or relief organization.

Emergency workers will be assisting people in flooded areas. You can help them by staying off the roads and out of the way.

Play it safe. Additional flooding or flash floods can occur. Listen for local warnings and information. If your car stalls in rapidly rising waters, get out immediately and climb to higher ground.

Return home only when authorities indicate it is safe.

Before:

• Build an emergency kit and make a family communications plan.
• Avoid building in a floodplain unless you elevate and reinforce your home.
• Elevate the furnace, water heater and electric panel in your home if you live in an area that has a high flood risk.
• Consider installing "check valves" to prevent flood water from backing up into the drains of your home.
• If feasible, construct barriers to stop floodwater from entering the building and seal walls in basements with waterproofing compounds.

During:

• Listen to the radio or television for information.
• Be aware that flash flooding can occur. If there is any possibility of a flash flood, move immediately to higher ground. Do not wait for instructions to move.
• Be aware of stream, drainage channels, canyons and other areas known to flood suddenly. Flash floods can occur in these areas with or without typical warnings such as rain clouds or heavy rain.

After:

• Avoid moving water.
• Stay away from damaged areas unless your assistance has been specifically requested by police, fire, or relief organization.
• Emergency workers will be assisting people in flooded areas. You can help them by staying off the roads and out of the way.
• Play it safe. Additional flooding or flash floods can occur. Listen for local warnings and information. If your car stalls in rapidly rising waters, get out immediately and climb to higher ground.
• Return home only when authorities indicate it is safe.

a: ΔABC vuông tại A

mà AM là đường trung tuyến

nên MA=MB=MC

Xét tứ giác AMCD có

AM//CD

CM//AD

Do đó: AMCD là hình bình hành

Hình bình hành AMCD có MA=MC

nên AMCD là hình thoi

b: Xét ΔAHI vuông tại H và ΔBAC vuông tại A có

\(\hat{HAI}=\hat{ABC}\) (hai góc đồng vị, AH//BC)

Do đó: ΔAHI~ΔBAC

a) Ta có: E,F lần lược là hình chiếu của B,C trên AD 

\(\Rightarrow\left\{{}\begin{matrix}\widehat{BEA}=90^o\\\widehat{CFA}=90^o\end{matrix}\right.\) 

Xét hai tam giác ABE và ACF có:

\(\widehat{BEA}=\widehat{CFA}\left(=90^o\right)\) 

\(\widehat{BAE}=\widehat{CAF}\) (do AD là phân giác của góc A) 

\(\Rightarrow\Delta ABE\sim\Delta ACF\left(g.g\right)\)  

b) Xét hai tam giác BDE và CDF có:

\(\widehat{BDE}=\widehat{CDF}\) (đối đỉnh) 

\(\widehat{BED}=\widehat{CFD}\left(=90^o\right)\)

\(\Rightarrow\Delta BDE\sim\Delta CDF\left(g.g\right)\) 

\(\Rightarrow\dfrac{BE}{CF}=\dfrac{DE}{DF}\) (1) 

Mà: \(\Delta ABE\sim\Delta ACF\left(cmt\right)\)

\(\Rightarrow\dfrac{BE}{CF}=\dfrac{AE}{AF}\left(2\right)\)

Từ (1) và (2) ta có: \(\dfrac{DE}{DF}=\dfrac{AE}{AF}\Rightarrow AF\cdot DE=AE\cdot DF\)