Ta có: \(\left(x^{3n}+y^{3n}\right)\left(x^{3n}-y^{3n}\right)=-x^6-y^6\)

\(\Leftrightarrow x^{6n}-y^{6n}=-x^6-y^6\)

\(\Leftrightarrow\left\{{}\begin{matrix}n=-1\\n=1\end{matrix}\right.\Leftrightarrow n\in\varnothing\)

f(0)=5

f(-3)=-2+5=3

1, \(2x^2+4x=2x\left(x+2\right)\)

2, \(15x^3+5x^2-10x=5x\left(3x^2+x-2\right)=5x\left(x-\dfrac{2}{3}\right)\left(x+1\right)\)

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

4) \(3\left(x-y\right)+5x\left(y-x\right)=\left(x-y\right)\left(3-5x\right)\)

5) \(5x^2-10x=5x\left(x-2\right)\)

6) \(3x-6y=3\left(x-2y\right)\)

7) \(25x^2+5x^3+x^2y=x^2\left(25+5x+y\right)\)

8) \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)

9) \(x\left(y-1\right)-y\left(y-1\right)=\left(x-1\right)\left(y-1\right)\)

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

\(1,=2x\left(x+2\right)\\ 2,=5x\left(3x^2+x-2\right)\\ 3,=\left(x-2y\right)\left(5x^2+15x\right)=5x\left(x+3\right)\left(x-2y\right)\\ 4,=\left(x-y\right)\left(3-5x\right)\\ 5,=5x\left(x-2\right)\\ 6,=3\left(x-2y\right)\\ 7,=5x^2\left(5+x+y\right)\\ 8,=7xy\left(2x-3y+4xy\right)\\ 9,=\left(y-1\right)\left(x-y\right)\\ 10,=\left(x-y\right)\left(10x+8y\right)=2\left(5x+4y\right)\left(x-y\right)\)

\(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}=\dfrac{2y-4}{6}=\dfrac{x-1+2y-4-z+2}{5+6-2}=\dfrac{6-5}{9}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}9x-9=5\\9y-18=3\\9z-18=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{9}\\y=\dfrac{7}{3}\\z=\dfrac{20}{9}\end{matrix}\right.\)

giúp mình với đi mà

\(a.y:\dfrac{2}{3}=\dfrac{4}{9}\times3\\y:\dfrac{2}{3}=\dfrac{4}{3}\\ y=\dfrac{4}{3}\times\dfrac{2}{3}\\ y=\dfrac{8}{9}\\ b.\dfrac{5}{6}+\dfrac{y}{12}=\dfrac{7}{6}\\ \dfrac{y}{12}=\dfrac{7}{6}-\dfrac{5}{6}\\ \dfrac{y}{12}=\dfrac{2}{6}\\ y\times6=12\times2=24\\ y=24:6\\ y=4 \)

\(a,y:\dfrac{2}{3}=\dfrac{4}{9}.3\)

\(y:\dfrac{2}{3}=\dfrac{4}{3}\)

\(y=\dfrac{4}{3}.\dfrac{2}{3}\)

\(y=\dfrac{8}{9}\)

\(b,\dfrac{5}{6}+\dfrac{y}{12}=\dfrac{7}{6}\)

\(\dfrac{y}{12}=\dfrac{7}{6}-\dfrac{5}{6}\)

\(\dfrac{y}{12}=\dfrac{2}{6}\)

\(y.6=12.2\)

\(y.6=24\)

\(y=24:6\)

\(y=4\)

\(\left(y-\dfrac{2}{3}\right):\dfrac{4}{5}=\dfrac{5}{6}\)

\(y-\dfrac{2}{3}=\dfrac{5}{6}\times\dfrac{4}{5}\)

\(y-\dfrac{2}{3}=\dfrac{2}{3}\)

\(y=\dfrac{2}{3}+\dfrac{2}{3}\)

\(y=\dfrac{4}{3}\)

\(y-\dfrac{2}{3}=\dfrac{5}{6}\times\dfrac{4}{5}\)

\(y-\dfrac{2}{3}=\dfrac{2}{3}\)

\(y=\dfrac{2}{3}+\dfrac{2}{3}\)

\(y=\dfrac{4}{3}\)

y – 2/3 = 5/6 x 4/5

y – 2/3 = 2/3

y= 2/3 + 2/3

y = 4/3