1139 là số nguyên tố.a\

Ta có: \(\frac{x}{-3}=\frac{y}{7}\Leftrightarrow\frac{x}{6}=\frac{y}{-14}\) mà \(\frac{y}{-14}=\frac{z}{5}\)

=> \(\frac{x}{6}=\frac{y}{-14}=\frac{z}{5}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có: 

\(\frac{x}{6}=\frac{y}{-14}=\frac{z}{5}=\frac{2x+4y-6z}{12-56-30}=-\frac{15}{74}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{15}{74}\cdot6=-\frac{45}{37}\\y=-\frac{15}{74}\cdot\left(-14\right)=\frac{105}{37}\\z=-\frac{15}{74}\cdot-\frac{75}{74}\end{cases}}\)

a) \(\left(-12x^{13}y^{15}+6x^{10}y^{14}\right):\left(-3x^{10}y^{14}\right)\)

\(=-12x^{13}y^{15}:-3x^{10}y^{14}+6x^{10}y^{14}:-3x^{10}y^{14}\)

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

b) \(\left(x-y\right)\left(x^2-2x+y\right)-x^3+x^2y\)

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

\(=-2x^2+3xy-y^2\) 

a) \(-12x^{13}\)\(y^{15}\)+\(6x^{10}\)\(y^{14}\):\(-3x^{10}\)\(y^{14}\)

=\(-12x\)\(^{13}\)\(y^{15}\)\(:\)\(-3x^{10}y^{14}\)\(+6x^{10}y^{14}:-3x^{10}y^{14}\)

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

b)\(=\left(x-y\right)x^2-2x+y-x^3+x^2y\)

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

\(=-2x+y\)

đầu cắt moi :)))))

11) \(3x\left(x-1\right)+5\left(1-x\right)=\left(3x-5\right)\left(x-1\right)\)

12) \(2\left(2x-1\right)+3\left(1-2x\right)=1-2x\)

13) \(10x\left(x-y\right)-8y\left(y-x\right)=2\left(x-y\right)\left(5x+4y\right)\)

14) \(3x\left(y+2\right)-3\left(y+2\right)=3\left(x-1\right)\left(y+2\right)\)

15) \(x^2-y^2-2x+2y\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)