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}

O m y x z

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

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

Đặt \(k=\dfrac{x}{2}=\dfrac{y}{2}=\dfrac{z}{-3}\)

\(\Rightarrow x=2k;y=2k;z=-3k\)

Ta thay \(x=2k;y=2k;z=-3k\) vào \(xyz=240\)

\(\Rightarrow\left(2k\right)\left(2k\right)\left(-3k\right)=240\)

\(\Rightarrow\left[2.2.\left(-3\right)\right]\left(k.k.k\right)=240\)

\(\Rightarrow\left(-12k^3\right)=240\)

\(\Rightarrow k=-\sqrt[3]{20}\)

Vậy \(\left\{{}\begin{matrix}x=y=-2\sqrt[3]{20}\\z=3\sqrt[3]{20}\end{matrix}\right.\)

\(x-y-z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y+z\\y=x-z\\z=x-y\end{matrix}\right.\) 

Lại có :\(C=\left(\dfrac{1-z}{x}\right)\left(\dfrac{1-x}{y}\right)\left(\dfrac{1+y}{z}\right)\)

\(C=\left(\dfrac{x-z}{x}\right)\left(\dfrac{y-x}{y}\right)\left(\dfrac{z+y}{z}\right)\)

\(C=\dfrac{y}{x}\cdot\) \(\left(-\dfrac{z}{y}\right)\) \(\dfrac{x}{z}\)

\(C=\dfrac{-xyz}{xyz}=-1\)