a: Gọi O là trung điểm của CD

=>O là tâm đường tròn đường kính CD

Xét (O) có

ΔCED nội tiếp

CD là đường kính

Do đó: ΔCED vuông tại E

=>\(\hat{CED}=90^0\)

Xét tứ giác ABCE có \(\hat{CAB}=\hat{CEB}=90^0\)

nên ABCE là tứ giác nội tiếp

b: EFDC là tứ giác nội tiếp

=>\(\hat{EFD}+\hat{ECD}=180^0\)

mà \(\hat{EFD}+\hat{AFD}=180^0\) (hai góc kề bù)

nên \(\hat{AFD}=\hat{ACE}\)

Xét ΔAFD và ΔACE có

\(\hat{AFD}=\hat{ACE}\)

góc FAD chung

Do đó: ΔAFD~ΔACE

=>\(\frac{AF}{AC}=\frac{AD}{AE}\)

=>\(AF\cdot AE=AD\cdot AC\)

a: \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{ab}\)

Ta có: \(\frac{\sqrt{a^3}-\sqrt{a^2b}+\sqrt{ab^2}-\sqrt{b^3}}{\sqrt{a}-\sqrt{b}}\)

\(=\frac{a\cdot\sqrt{a}-a\cdot\sqrt{b}+b\cdot\sqrt{a}-b\cdot\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

\(=\frac{a\left(\sqrt{a}-\sqrt{b}\right)+b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}=a+b\)

Ta có: \(P=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{a^2b}+\sqrt{ab^2}-\sqrt{b^3}}{\sqrt{a}-\sqrt{b}}\)

\(=\sqrt{ab}-a-b\)

b: \(P=-a+\sqrt{ab}-b\)

\(=-\left(a-\sqrt{ab}+b\right)=-\left(a-\sqrt{ab}+\frac14b+\frac34b\right)\)

\(=-\left(\sqrt{a}-\frac12\sqrt{b}\right)^2-\frac34b<0\forall a,b\) thỏa mãn ĐKXĐ

c: \(P=2\sqrt{ab}-b\)

=>\(\sqrt{ab}-a-b=2\sqrt{ab}-b\)

=>\(-a-\sqrt{ab}=0\)

=>\(a+\sqrt{ab}=0\)

=>\(\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)=0\) (vô lý vì a>0; b>0)

=>(a;b)∈∅

Gọi Kim loại Đó là A

\(A+2HCl\rightarrow ACl_2+H_2\)

tl1.......2.............1...........1(mol)

br  0,15....0,3.....0,15.....0,15(mol)                                 

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(M_A=\dfrac{m}{n}=\dfrac{3,6}{0,15}=24\left(\dfrac{g}{mol}\right)\)

Vậy A là Magie(Mg)

UHH THÌ BẠN ĐANG ĐĂNG MÀ

cosplay chim cánh cụt đi

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

em cám ơn thầy nhiều ạ!

26.

\(\sqrt{\dfrac{-3}{2a^3}}=\sqrt{\dfrac{-3a}{2a^4}}=\dfrac{1}{a^2}\sqrt{\dfrac{-3a}{2}}\)

Đáp án B

28.

\(\sqrt{\dfrac{a^3}{a}}=\sqrt{a^2}=\left|a\right|=-a\)

Đáp án B

Hello! John.
Glad to see you again. Tam, this is my cousin, Peter. It's his first to visit to your country.
How do you do? Welcome to VN
Thank you. Nice to meet you, Tam
Can I help you with your suitcases, John?
Thanks. I can manage.
OK. Now we're going to the hotel in the center of the city by taxi
That would be nice. How far is it from the airport to the center of the city?
It's about a half-hour drive
Look, John! What a lot of motorbikes in the streets!
Oh, yeah. That surprised me by the time I first came to VN
Motorbikes are our main means of transport. I go to school every day by bike.
Would you mind taking me around the city by bike?
No, of course not. But I have to ask someone else to get Peter, too
Peter, do you mind if Tam's friend gives us a ride around the city?
No, I don't mind. But I feel a little bit scared because the traffic is so heavy

OK. So we'll go on a sightseeing tour by bikes at weekend

Câu 4:

4.1/ Ta có: \(n_{NaCl}=2,5.0,4=1\left(mol\right)\)

\(\Rightarrow m_{NaCl}=1.58,5=58,5\left(g\right)\)

4.2/ Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

_____0,2___________0,2____0,2 (mol)

b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

Bạn tham khảo nhé!

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