bài nào vậy bn:?

a)Đk \(x\ge0,x\ne1\)

\(\Rightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

\(\Rightarrow P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2-\sqrt{3}-1}{2-\sqrt{3}+1}=\dfrac{1-\sqrt{3}}{3-\sqrt{3}}=\dfrac{-\sqrt{3}}{3}\)

c)\(TacoP>\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}>\dfrac{1}{2}\)

\(\Leftrightarrow2\sqrt{x}-2>\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)

Vậy\(x>9\)Thì \(P>\dfrac{1}{2}\)

Bài 3.37

a) Do BE // AC (gt)

⇒ ∠ABE = ∠A = 80⁰ (so le trong)

Do CF // AB

⇒ ∠ACF = ∠A = 80⁰ (so le trong)

⇒ ∠ABE = ∠ACF = 80⁰

b) Do CF // AB

⇒ ∠FCz = ∠ABC = 60⁰ (đồng vị)

Ta có:

∠BCF + ∠FCz = 180⁰ (kề bù)

⇒ ∠BCF = 180⁰ - ∠FCz

= 180⁰ - 60⁰

= 120⁰

⇒ ∠ACB = ∠BCF - ∠ACF

= 120⁰ - 80⁰

= 40⁰

c) Do Bx là tia phân giác của ∠ABE

⇒ ∠ABx = ∠ABE : 2 = 80⁰ : 2 = 40⁰

⇒ ∠xBC = ∠ABx + ∠ABC

= 40⁰ + 60⁰

= 100⁰

Do Cy là tia phân giác của ∠ACF

⇒ ∠yCF = ∠ACF : 2 = 80⁰ : 2 = 40⁰

⇒ ∠yCz = ∠yCF + ∠FCz

= 40⁰ + 60⁰

= 100⁰

⇒ ∠xBC = ∠yCz = 100⁰

Mà ∠xBC và ∠yCz là hai góc đồng vị

⇒ Bx // Cy

Bài 3.36

a) Do Ox // AB

⇒ ∠BOx = ∠ABO = 40⁰ (so le trong)

b) Ta có:

∠xOD = ∠BOD - ∠BOx

= 110⁰ - 40⁰

= 70⁰

⇒ ∠xOD = ∠ODC = 70⁰

Mà ∠xOD và ∠ODC là hai góc so le trong

⇒ Ox // CD

Mà Ox // AB (gt)

⇒ AB // CD

1)
\(\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}\right]:\dfrac{2\sqrt{3x}}{x-1}\)
\(=\left(\dfrac{x+\sqrt{x}+x-\sqrt{x}}{x-1}\right).\dfrac{x-1}{2\sqrt{3x}}\)
\(=\dfrac{2x}{x-1}.\dfrac{x-1}{2\sqrt{3x}}=\dfrac{\sqrt{x}}{\sqrt{3}}=\dfrac{\sqrt{3x}}{3}\)

1a  2b   3b

1.A

2.B

3.B

1.A

2.B

3.B

a: MNPQ là hình bình hành

=>MQ//NP

=>MQ//IP

Xét tứ giác MIPQ có IP//MQ

nên MIPQ là hình thang

b: ΔMNP vuông cân tại N

=>MN=NP và \(\widehat{MNP}=90^0\)

Hình bình hành MNPQ có \(\widehat{MNP}=90^0\)

nên MNPQ là hình chữ nhật

=>\(\widehat{Q}=\widehat{P}=90^0\)

Xét ΔMNI vuông tại N có \(sinNMI=\dfrac{NI}{MN}=\dfrac{2}{3}\)

nên \(\widehat{NMI}\simeq42^0\)

\(\widehat{NMI}+\widehat{QMI}=\widehat{NMQ}=90^0\)

=>\(\widehat{QMI}+42^0=90^0\)

=>\(\widehat{QMI}=48^0\)

IP//MQ

=>\(\widehat{QMI}+\widehat{MIP}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{MIP}+48^0=180^0\)

=>\(\widehat{MIP}=132^0\)

1. Going in the rain is interesting.

2. Don't forget to learn by heart new words.

3. If he doesn't apologize me, I won't forgive him.

4. My grandmother is the most helpful (person) in my village.

5. What is the price of that hat?

6. Your book is different from my book.

1 Her mother is not as hospitable as her

2 Hung is fond of swimming in the summer

3 Let's travel abroad

4 There are 4 chairs in my room

5 Don't forget to drink more water

6 Minh is the most intelligent student in my class

$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$n_{H_2} = n_{Zn} = \dfrac{19,5}{65} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH : 

$n_{Al} = \dfrac{2}{3}n_{H_2} = 0,2(mol)$
$m_{Al} = 0,2.27 = 5,4(gam)$