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III

1 It's colder today than yesterday

2 It takes 4 hours to travel by car and fives hours by train

3 We were busier at work today than everyday

4 Jane's sister cooks worse than her

5 Nobody in this team can play football as well as Tom

IV

1 D
2 A

e thay dấu = cho tất cả phsố trog bài 3 rồi tìm x , khi tìm x thì coi dấu của bài r nói x lớn hoặc nhỏ hơn số đó là đc

4:

a: =>4x^4-4x^2+x^2-1=0

=>(x^2-1)(4x^2+1)=0

=>x^2-1=0

=>x=1 hoặc x=-1

b: ĐKXĐ: x<>5; x<>2

PT =>\(\dfrac{x-2}{x-5}+3=\dfrac{6}{x-2}\)

=>\(x^2-4x+4+3\left(x^2-7x+10\right)=6x-30\)

=>4x^2-25x+34-6x+30=0

=>4x^2-31x+64=0

=>\(x\in\varnothing\)

c: =>x^2(2x^2+5)+2=0

=>x^2(2x^2+5)=-2(vôlý)

d: =>(2x-5)(x-2)=3x(x-1)

=>3x^2-3x=2x^2-4x-5x+10

=>x^2+6x-10=0

=>\(x=-3\pm\sqrt{19}\)

e: ĐKXĐ: x<>3; x<>-2

PT =>x^2-3x+5=x+2

=>x^2-4x+3=0

=>(x-3)(x-1)=0

=>x=1(nhận) hoặc x=3(loại)

f: ĐKXĐ: x<>2; x<>3

PT =>2x(x-3)-5(x-2)=5

=>2x^2-6x-5x+10-5=0

=>2x^2-11x+5=0

=>2x^2-10x-x+5=0

=>(x-5)(2x-1)=0

=>x=1/2 hoặc x=5

Bài 3: 

PTHH: \(2R+O_2\xrightarrow[]{t^O}2RO\)

Theo PTHH: \(n_R=n_{RO}\) \(\Rightarrow\dfrac{3,6}{M_R}=\dfrac{6}{M_R+16}\)

  \(\Rightarrow M_R=24\)  (Magie)

Bài 4 ở đâu vậy ??

Bài 2: 

 PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,9\left(mol\right)\\n_{AlCl_3}=0,3\left(mol\right)\\n_{H_2}=0,45\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,45\cdot22,4=10,08\left(l\right)\\m_{AlCl_3}=0,3\cdot133,5=40,05\left(g\right)\\m_{ddHCl}=\dfrac{0,9\cdot36,5}{7,3\%}=450\left(g\right)\\m_{H_2}=0,45\cdot2=0,9\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Al}+m_{ddHCl}-m_{H_2}=457,2\left(g\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{40,05}{457,2}\cdot100\%\approx8,76\%\) 

 Bài 2 : 

$n_{Ba(OH)_2} = 0,3(mol) ; n_{BaSO_3} = 0,08(mol)$

TH1 : $Ba(OH)_2$ dư

$Ba(OH)_2 + SO_2 \to BaSO_3 + H_2O$
$n_{SO_2} = n_{BaSO_3} = 0,08(mol)$
$V_{SO_2} = 0,08.22,4 = 1,792(lít)$

TH2 : có tạo muối axit

SO₂ + Ba(OH)₂ → BaSO₃ + H₂O

0,08......0,08..............0,08..............(mol)

2SO₂ + Ba(OH)₂ → Ba(HSO₃)₂

0,44........0,22......................................(mol)

$V_{SO_2} = (0,08 + 0,44).22,4 = 11,648(lít)$

Bài 4 : 

$n_{BaCO_3} = 0,05(mol)$

CO₂ + Ba(OH)₂ → BaCO₃ + H₂O

0,05......0,05..............0,05..............(mol)

2CO₂ + Ba(OH)₂ → Ba(HCO₃)₂

..............0,2..................0,2....................(mol)

Ba(HCO₃)₂ \(\xrightarrow{t^o}\)BaCO₃ + CO₂ + H₂O

0,2.....................0,2........................(mol)

m = 0,2.197 = 39,4 gam

4. \(n_{BaCO_3}=\dfrac{9,85}{197}=0,05\left(mol\right)\)

Bảo toàn nguyên tố Ba =>  \(n_{Ba\left(HCO_3\right)_2}=0,25-0,05=0,2\left(mol\right)\)

Lọc kết tủa đem nung thì thu được m gam chất rắn : BaCO3 

Bảo toàn nguyên tố Ba =>\(n_{BaCO_3}=n_{Ba\left(HCO_3\right)_2}=0,2\left(mol\right)\)

=> \(m=0,2.197=39,4\left(g\right)\)

2b)

Áp dụng BĐT bunhiacopxki có:

\(\left(1+1\right)\left(x^4+y^4\right)\ge\left(x^2+y^2\right)^2\)

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2\)\(\Leftrightarrow x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)

\(\Rightarrow2\left(x^4+y^4\right)\ge\dfrac{\left(x+y\right)^4}{4}\Leftrightarrow x^4+y^4\ge\dfrac{1}{8}.\left(x+y\right)^4\)

Dấu "=" xảy ra khi x=y

3)

Áp dụng bđt Holder có:

\(\left(x^3+y^3+z^3\right)\left(1+1+1\right)\left(1+1+1\right)\ge\left(x+y+z\right)^3\)

\(\Leftrightarrow x^3+y^3+z^3\ge\dfrac{1}{9}\left(x+y+z\right)^3\)

Dấu "=" xảy ra khi x=y=z

3)(Nếu không dùng Holder)

Với x,y,z >0, ta có bđt sau:\(2x^3+2y^3+2z^3\ge xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\) (1)

Thật vậy (1)\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)+\left(y+z\right)\left(y^2-yz+z^2\right)-yz\left(y+z\right)+\left(z+x\right)\left(z^2-zx+x^2\right)-zx\left(x+z\right)\ge0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)^2+\left(y+z\right)\left(y-z\right)^2+\left(z+x\right)\left(z-x\right)^2\ge0\) (lđ)

Áp dụng AM-GM có:

\(x^3+y^3+z^3\ge3xyz\)

\(\Leftrightarrow\dfrac{2\left(x^3+y^3+z^3\right)}{3}\ge2xyz\) (2)

Từ (1) và (2), cộng vế với vế \(\Rightarrow\dfrac{8}{3}\left(x^3+y^3+z^3\right)\ge xy\left(x+y\right)+yz\left(x+z\right)+xz\left(x+z\right)+2xyz\)

\(\Leftrightarrow\dfrac{8}{3}\left(x^3+y^3+z^3\right)\ge\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(\Leftrightarrow9\left(x^3+y^3+z^3\right)\ge x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)^3\)

\(\Rightarrow x^3+y^3+z^3\ge\dfrac{1}{9}\left(x+y+z\right)^3\) (đpcm)