13y + y - 4y - 5 = 1015

13 . y + y . 1 - 4 . y - 5 = 1015

y . ( 13 + 1 - 4 ) - 5 = 1015

y . 10 - 5 = 1015

y . 10 = 1015 + 5

y . 10 = 1020

y = 1020 : 10

ý = 102

13y + y - 4y -5 = 1015

13y + y - 4y = 1015 + 5 = 1020

10y = 1020

=> y = 1020 : 10 = 102

a)Ta có 6x=4y=-2z và x-y-z=27

\(\Rightarrow6x.\dfrac{1}{12}=4y.\dfrac{1}{12}=-2z.\dfrac{1}{12}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-6}=\dfrac{x-y-z}{2-3-\left(-6\right)}=\dfrac{27}{5}\)

\(\Rightarrow x=2.\dfrac{27}{5}=10,8\)

\(y=3.\dfrac{27}{5}=16,2\)

\(\Rightarrow z=-6.\dfrac{27}{5}=-32,4\)

b) Ta có 13y=6z

\(\Rightarrow\dfrac{y}{6}=\dfrac{z}{13}\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{13}\) và x.y.z=576(1)

Đặt \(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{13}=k\Rightarrow x=4k;y=6k;z=13k\)(2)

Thay (2) vào (1) ta được

\(4k.6k.13k=576\)

\(\Rightarrow312.k^3=576\)

mk làm tới đây thì chia k đc

a) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)

\(\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}15y-3=0\\4y-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{5}\\y=\frac{9}{4}\end{matrix}\right.\)

Vây \(y\in\left\{\frac{1}{5};\frac{9}{4}\right\}\)

b) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)

\(\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}z=\frac{8}{27}\\z=-\frac{7}{25}\end{matrix}\right.\)

Vậy \(z\in\left\{\frac{8}{27};-\frac{7}{25}\right\}\)

c) \(13y\left(y-8\right)-2y+16=0\)

\(\Leftrightarrow13y\left(y-8\right)-2\left(y-8\right)=0\)

\(\Leftrightarrow\left(13y-2\right)\left(y-8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{2}{13}\\y=8\end{matrix}\right.\)

Vậy \(y\in\left\{\frac{2}{13};8\right\}\)

d) \(-10y\left(y+2\right)-y-2=0\)

\(\Leftrightarrow\left(-10y-1\right)\left(y+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-\frac{1}{10}\end{matrix}\right.\)

Vậy \(y\in\left\{-2;-\frac{1}{10}\right\}\)

e) \(x\left(x+19\right)^2-\left(x+19\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+19\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-19\end{matrix}\right.\)

Vậy \(x\in\left\{1;-19\right\}\)