Giúp mình với, mình cần gấp

Ta có \(\dfrac{1}{12};\dfrac{1}{13};..;\dfrac{1}{19}< \dfrac{1}{11}\)

=> \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..+\dfrac{1}{19}< \dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+..+\dfrac{1}{11}\) (9 số hạng )

\(C< 9\cdot\dfrac{1}{11}=\dfrac{9}{11}< 1\) (1)

Vì \(\dfrac{1}{11};\dfrac{1}{12};\dfrac{1}{13};...;\dfrac{1}{19}>0\) => C>0 (2)

Từ (1);(2) có 0<C<1

=> C không phải là số nguyên 

`(3 5/6 - 1 1/3) . ( 3 14/15 - 2 3/5)`

`= ( 23/6 - 4/3) . (59/15 - 13/5)`

`= ( 23/6 - 8/6) . (59/15 - 39/15)`

`=((23-8)/6) . ((59-39)/15)`

`= 15/6 . 20/15`

`=5/2 . 4/3`

`=(5.4)/(2.3)`

`=20/6`

`=10/3`

Mình không hiểu lắm =(

(x+3)⋮ x ⇔ 3 ⋮ x ⇔ x ϵ Ư(3) ⇔ x ϵ {-1; -3; 1; 3}

x + 7 ⋮ x + 3 ⇔ x + 3 + 4 ⋮ x + 3 ⇔ 4 ⋮ x + 3 

⇔ x + 3 ϵ Ư(4) ⇔ x ϵ { -7; -5; -4 -2; -1; 1}

c, 3x -5 ⋮ x - 1 ⇔ 3(x-1) - 2 ⋮x -1 ⇔ 2 ⋮ x-1

⇔x- 1 ϵ Ư(2) ⇔ x ϵ {-1; 0; 2; 3}

Theo đề ra: Tia Oz là tia phân giác \(\widehat{xOy}\)

\(\Rightarrow\widehat{xOz}=\widehat{xOy}:2=60^o:2=30^o\)

a, \(M\left(x\right)+N\left(x\right)=\) \(\left(-6x^2-7+2x+5x\right)+\left(12+6x^2-4x-3x\right)\)

\(M\left(x\right)+N\left(x\right)=-6x^2-7+2x+5x+12+6x^2-4x-3x\)

\(M\left(x\right)+N\left(x\right)=\left(-6x^2+6x^2\right)+\left(2x+5x-4x-3x\right)+\left(-7+12\right)\)

\(M\left(x\right)+N\left(x\right)=5\)

b, \(M\left(x\right)-N\left(x\right)=\left(-6x^2-7+2x+5x\right)-\left(12+6x^2-3x-4x\right)\)

\(M\left(x\right)-N\left(x\right)=-6x^2-7+2x+5x-12-6x^2+3x+4x\)

\(M\left(x\right)+N\left(x\right)=\left(-6x^2-6x^2\right)+\left(2x+5x+3x+4x\right)+\left(-7-12\right)\)

\(M\left(x\right)+N\left(x\right)=-12x^2+14x-19\)

c, \(P\left(x\right)=N\left(x\right)+4x3+3x-12\)

\(P\left(x\right)=\left(12+6x^2-4x-3x\right)+12x+3x-12\)

\(P\left(x\right)=12+6x^2-4x-3x-12x+3x-12\)

\(P\left(x\right)=6x^2+\left(-3x-4x+12x+3x\right)+\left(12-12\right)\)

\(P=6x^2+8x\)

Bậc của đa thức:2

Hệ số cao nhất :6

Hệ số tự do :0

a)3/7x(-8/2)+(-3/5)= -2/7 + (-3/5)= -31/35

b)(4/3)+(-2/5)+(-3/2)= 14/15 + (-3/2)= -17/30

c)4/5-(-2/7)- 7/10 =38/35 - 7/10 = 27/70

\(\dfrac{9^9+27^7}{9^6+243^3}=\dfrac{\left(3^2\right)^9+\left(3^3\right)^7}{\left(3^2\right)^6+\left(3^5\right)^3}=\dfrac{3^{18}+3^{21}}{3^{12}+3^{15}}=\dfrac{3^{18}\left(1+3^3\right)}{3^{12}\left(1+3^3\right)}\) \(=\dfrac{3^{18}}{3^{12}}=3^6\)

(-3/4)^2= 9/16

\((-\dfrac{3}{4})^2=\dfrac{9}{16}\)