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tìm số hữu tỉ a và b

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a=2b 

=>a:b=2

=>4(a+b)=2

=> a+b=1/2

=>3b=1/2

=>b=1/6;a =2b=1/3

\(\left(2x-3y\right).5xy\\ =5xy.2x+5xy.\left(-3y\right)\\ =10x^2y-15xy^2\)

\(\dfrac{x+2022}{2020}+\dfrac{x-2016}{2018}=\dfrac{x+2021}{2019}+\dfrac{x-2019}{2021}\\ \Rightarrow\left(\dfrac{x+2022}{2020}-1\right)+\left(\dfrac{x-2016}{2018}+1\right)=\left(\dfrac{x+2021}{2019}-1\right)+\left(\dfrac{x-2019}{2021}+1\right)\\ \Rightarrow\dfrac{x+2}{2020}+\dfrac{x+2}{2018}-\dfrac{x+2}{2019}-\dfrac{x+2}{2021}=0\\ \Rightarrow\left(x+2\right)\left(\dfrac{1}{2020}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2021}\right)=0\\ \)

\(\Rightarrow x+2=0\) ( Vì: \(\dfrac{1}{2020}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2021}>0\) )

\(\Rightarrow x=-2\)

Everyone has their own origin, and each individual's could be coming from many angles, but one of the most important aspects to determind is where they got raised - their hometown. Mine is Hanoi!

I was a normal kid born in Hanoi City. It's iconic capital of Vietnam, known for its wholesome cuisine, elegant aesthetic and friendliness. Hanoi is a beautiful place where you can encounter many foreginers crossing by. Besides admiring the scenery, you may also find yourself immersed in various kinds of activities such as doing excercise, skating or going for simple a walk, which is some kind of healthy lives, isn't it? Hanoi's beauty is something I really love and really appreciate.

Loving your hometown is a thing but knowing how to improve and make your place a better one way more worthy. So I'll try my best to protect and make the captial of Vietnam, my beloved hometown a even more remarkable and a even more lovely place in the future.

1: Sửa đề: \(f\left(x\right)=3x\left(1-3x+2x^3\right)-2x^2\left(-4+3x^2-x\right)\)

\(=3x-9x^2+6x^4+8x^2-6x^4+2x^3\)

\(=2x^3-x^2+3x\)

\(g\left(x\right)=-4\left(x^4+x^2+1\right)+x^3\left(4x+2\right)+4\)

\(=-4x^4-4x^2-4+4x^3+2x^3+4\)

\(=2x^3-4x^2\)

Bậc là 3

Hệ số cao nhất là 2

Hệ số tự do là 0

2: f(x)=g(x)+h(x)

=>h(x)=f(x)-g(x)

\(=2x^3-x^2+3x-2x^3+4x^2=3x^2+3x\)

3: Đặt h(x)=0

=>3x(x+1)=0

=>x(x+1)=0

=>\(\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

1. `G(x)=-4(x^4+x^2+1)+x^3(4x+2)+4`

`=-4x^4-4x^2-4+4x^4+2x^3+4`

`=(4x^4-4x^4)+2x^3-4x^2+(4-4)`

`=2x^3-4x^2`

Bậc 3

Hệ số cao nhất: 2

Hệ số tự đó: 0

2. `F(x) = G(x) + H(x)`

`=>H(x)=F(x) - G(x)`

`=>H(x)=[3x(1-3x+2x^3)-2x^2(-4+3x^2-x)]-(2x^3-4x^2)

`=>H(x)=3x-9x^2+6x^4+8x^2-6x^4+2x^3-2x^3+4x^2`

`=>H(x)=3x^2+3x`

3. `H(x)=3x^2+3x=0`

`=>3x(x+1)=0`

TH1: `x=0`

TH2: `x+1=0=>x=-1`

BÀI 1: 

a) \(\dfrac{9}{70}>\dfrac{5}{42}\)              b) \(-\dfrac{4}{27}>-\dfrac{10}{63}\)

c) \(\dfrac{999}{556}>\dfrac{1000}{557}\)        d) \(-\dfrac{2}{15}< \dfrac{10}{11}\)

BÀI 2:

\(-1\dfrac{2}{3}< -\dfrac{3}{5}< -\dfrac{5}{9}< 0,5< \dfrac{10}{9}\)

BÀI 3:

a) -3 ϵ Q     b) 10 ϵ N

c) -3/7 ϵ Q     d) -2 ϵ Q

`(-3^x*3^6)/(-27*9^x) = -3`

`(-3^(x+6))/(-3^3*(3^2)^x)=-3`

`(-3^(x+6))/(-3^3*3^(2x))=-3`

`(-3^(x+6))/(-3^(2x+3))=-3`

`3^(x+6-2x-3)=-3`

`3^(3-x)=-3`

`(3^(3-x))/3=-1`

`3^(3-x-1)=-1`

`3^(2-x)=-1`

Vì `3^(2-x) > 0` mà `-1<0`

Nên không có x thoả mãn

Bạn vt lại gt lại \(a\left(a+b\right)=c^2\) bạn áp dụng bổ đề trên ta được a+a+b+c+c là hợp số

\(\dfrac{\left(-\dfrac{1}{3}\right)^6}{\left(\dfrac{1}{6}\right)^2}=\dfrac{\left(\dfrac{1}{3}\right)^6}{\left(\dfrac{1}{6}\right)^2}\\ =\dfrac{\dfrac{1^6}{3^6}}{\dfrac{1^2}{6^2}}=\dfrac{\dfrac{1}{3^6}}{\dfrac{1}{6^2}}\\ =\dfrac{1}{3^6}:\dfrac{1}{6^2}=\dfrac{1}{3^6}.\dfrac{6^2}{1}\\ =\dfrac{6^2}{3^6}=\dfrac{2^2.3^2}{3^2.3^4}\\ =\dfrac{2^2}{3^4}=\dfrac{4}{81}\)

(-1/6)^3/(1/6)^2

= -1/18 / 1/36

= -1/18 ÷ 1/36

= -1/18 × 36/1

= -1/1 × 2/1

= -2