Câu 45: B

Câu 45:B

B

$M_O = 16(đvC)$

$PTK = Y + 16.2 = M_{Cu} = 64(đvC) \Rightarrow Y = 32$
Vậy Y là lưu huỳnh, KHHH : S

d) Ta có: \(1,25\cdot\left(-2\right)^2-0.5:\dfrac{5}{12}+15\%\)

\(=\dfrac{5}{4}\cdot4-\dfrac{1}{2}\cdot\dfrac{12}{5}+\dfrac{3}{20}\)

\(=5-\dfrac{6}{5}+\dfrac{3}{20}\)

\(=\dfrac{100}{20}-\dfrac{120}{20}+\dfrac{3}{20}\)

\(=\dfrac{-17}{20}\)

More progress in gender equality will be made by the Vietnamese government

Old men should be treated with respect by children

2 câu dưới mình làm rồi nha bạn =))

Thi xong nên giúp nek :)))

ủa thi?

1.Jackson is from China 

2.He listens to music in his free time 

3.She is a dancer

4.Yes, she does 

5.Tung is from Vietnam

6.He sufts the internet in his free time 

1. Jackson is from china

2.Jackson listens to music in his free time

3.Lisa is a dancer

4.Yes she does

5.Tung is from Vietnam

6.Tung surfs internet in his free time.

"Chúc em học tốt"

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)

\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

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

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

b: \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\)

Khi \(x=\left(2-\sqrt{3}\right)^2\) thì 

\(P=\dfrac{2}{\left(2-\sqrt{3}\right)^2+\sqrt{\left(2-\sqrt{3}\right)^2}+1}\)

\(P=\dfrac{2}{7-4\sqrt{3}+2-\sqrt{3}+1}\)

\(=\dfrac{2}{10-5\sqrt{3}}=\dfrac{4+2\sqrt{3}}{5}\)

c: P>=2/3

=>P-2/3>=0

=>\(\dfrac{2}{x+\sqrt{x}+1}-\dfrac{2}{3}>=0\)

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

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

=>\(-x-\sqrt{x}+2>=0\)

=>\(x+\sqrt{x}-2< =0\)

=>\(\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)< =0\)

=>\(\sqrt{x}-1< =0\)

=>0<=x<=1

Kết hợp ĐKXĐ, ta được: 0<=x<1