nSO3=0,75(mol)

mH2SO4(bđ)= 600.15%=90(g) => nH2SO4(bđ)= 90/98=45/49(mol)

PTHH: SO3 + H2O -> H2SO4

0,75______0,75____0,75(mol)

=> mH2SO4(tổng)= 0,75.98+90= 163,5(g)

mddH2SO4= mSO3+mddH2SO4(bđ)= 0,75.80 + 600= 660(g)

=>C%ddH2SO4(sau)= (163,5/660).100=24,773%

VddH2SO4(sau)= 660/1,25= 528(ml)=0,528(l)

nH2SO4(tổng)= 45/49 + 0,75= 327/196(mol)

=> CMddH2SO4(sau)= 327/196 : 0,528=3,16(M)

ĐK: \(x>0;x\ne1\)

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

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

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

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

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

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

\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}+2\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}+2=4\)

Đẳng thức không xảy ra nên \(C>4\).

C

CÂU C nha!

NhỚ tick mik nha! yÊu =)

4 C

5 A

6 B

7 D

8 C

9 A

10 B

Ex7

1 My brother is earning more than he was in his previous job

2 Beef is more expensive than Chicken

3 To my knowledge, listening to conversations in English is harder than speaking English

4 She has never had a more delicious lunch than this

5 Mr Long is the best doctor in this hospital

6 I have never driven a faster moving car than this

7 This castle is the oldest 

8 Is that the biggest shirt you have got?

9 I am the lowest one in the class

10 The company's revenue in the previous months were lower than in May

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=1\\x+\dfrac{1}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

bạn vẽ nốt hình đi để mình giúp nhé 

a: Xét tứ giác ABMC có 

D là trung điểm của AM

D là trung điểm của BC

Do đó: ABMC là hình bình hành

Suy ra: AC//BM và AC=BM

b: Xét ΔABM và ΔMCA có

AB=MC

BM=CA
AM chung

Do đó: ΔABM=ΔMCA

c: Xét ΔBHA vuông tại H và ΔCKM vuông tại K có

BA=CM

\(\widehat{ABH}=\widehat{MCK}\)

Do đó: ΔBHA=ΔCKM

Suy ra: BH=CK