Tính x,y,z
Biết x,y+3k-y=6
giúp mình vs mn ạ
nha nha
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Ta có: \(\left(x^{3n}+y^{3n}\right)\left(x^{3n}-y^{3n}\right)=-x^6-y^6\)
\(\Leftrightarrow x^{6n}-y^{6n}=-x^6-y^6\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=-1\\n=1\end{matrix}\right.\Leftrightarrow n\in\varnothing\)
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) \(10x\left(x-y\right)-8y\left(y-x\right)=\left(10x+8y\right)\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)
\(1,=2x\left(x+2\right)\\ 2,=5x\left(3x^2+x-2\right)\\ 3,=\left(x-2y\right)\left(5x^2+15x\right)=5x\left(x+3\right)\left(x-2y\right)\\ 4,=\left(x-y\right)\left(3-5x\right)\\ 5,=5x\left(x-2\right)\\ 6,=3\left(x-2y\right)\\ 7,=5x^2\left(5+x+y\right)\\ 8,=7xy\left(2x-3y+4xy\right)\\ 9,=\left(y-1\right)\left(x-y\right)\\ 10,=\left(x-y\right)\left(10x+8y\right)=2\left(5x+4y\right)\left(x-y\right)\)
\(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}=\dfrac{2y-4}{6}=\dfrac{x-1+2y-4-z+2}{5+6-2}=\dfrac{6-5}{9}=\dfrac{1}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}9x-9=5\\9y-18=3\\9z-18=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{9}\\y=\dfrac{7}{3}\\z=\dfrac{20}{9}\end{matrix}\right.\)
\(\left(y-\dfrac{2}{3}\right):\dfrac{4}{5}=\dfrac{5}{6}\)
\(y-\dfrac{2}{3}=\dfrac{5}{6}\times\dfrac{4}{5}\)
\(y-\dfrac{2}{3}=\dfrac{2}{3}\)
\(y=\dfrac{2}{3}+\dfrac{2}{3}\)
\(y=\dfrac{4}{3}\)