\(xy^2-9x=x.\left(y^2-3^2\right)=x.\left(y-3\right)\left(y+3\right)\)

\(x^2+14x+49-y^2=\left(x^2+2.7x+7^2\right)-y^2=\left(x+7\right)^2-y^2=\left(x+7-y\right).\left(x+7+y\right)\)

\(xy-y^2-x+y=y.\left(x-y\right)-\left(x-y\right)=\left(x-y\right).\left(y-1\right)\)

\(5x.\left(x-7\right)-x+7=5x.\left(x-7\right)-\left(x-7\right)=\left(x-7\right).\left(5x-1\right)\)

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

\(5x^3-40=5.\left(x^3-2^3\right)=5.\left(x-2\right).\left(x^2+2x+4\right)\)

\(x^2-y^2+12y-36=\left(y^2-2.6y+6^2-x^2\right)=-\left[\left(y-6\right)^2-x^2\right]\)\(=-\left[y-6-x\right].\left[y-6+x\right]\)

\(x^2z+4xyz+4y^2z=z.\left[x^2+2.2xy+\left(2y\right)^2\right]=z.\left(x+2y\right)^2\)

kudo shinichi cậu là cứu thế của mình

Bài 3 :

\(x=3y=2z\)

\(\Rightarrow x=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{2}}\)

\(\Rightarrow\frac{2x}{2}=\frac{3y}{1}=\frac{4z}{2}=\frac{2x-3y+4z}{2-1+2}=\frac{k}{3}\)

\(\Rightarrow x=\frac{k}{3}\)

     \(y=\frac{k}{3}.\frac{1}{3}=\frac{k}{9}\)

     \(z=\frac{k}{3}.\frac{1}{2}=\frac{k}{6}\)

a. \(x^2+4y^2+z^2=2x+12y-4z-14\)

\(\Leftrightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)

Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(2y-3\right)^2\ge0\\\left(z+2\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2y-3=0\\z+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

b. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)

Ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) => \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\)

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

   \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}=\frac{x^2+y^2-2z^2}{4+9-32}=\frac{76}{-19}=-4\)

=> \(\hept{\begin{cases}\frac{x^2}{4}=-4\\\frac{y^2}{9}=-4\\\frac{2z^2}{32}=-4\end{cases}}\) => \(\hept{\begin{cases}x^2=-4.4=-16\\y^2=-4.9=-36\\z^2=\left(-4.32\right):2=-64\end{cases}}\) => ko có giá trị x,y,z thõa mãn

Ta có: \(-2x=5y\) => \(\frac{x}{5}=\frac{y}{-2}\)

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

        \(\frac{x}{5}=\frac{y}{-2}=\frac{x+y}{5-2}=\frac{30}{3}=10\)

=> \(\hept{\begin{cases}\frac{x}{5}=10\\\frac{y}{-2}=10\end{cases}}\) => \(\hept{\begin{cases}x=10.5=50\\y=10.\left(-2\right)=-20\end{cases}}\)

Vậy ..

\(\frac{x}{-3}=\frac{y}{-7}\Rightarrow\frac{2x}{-6}=\frac{4y}{-28}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{2x}{-6}=\frac{4y}{-28}=\frac{2x+4y}{(-6)+(-28)}=\frac{68}{-34}=-2\)

Vậy : \(\hept{\begin{cases}\frac{x}{-3}=-2\\\frac{y}{-7}=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=14\end{cases}}\)

thanks hai bạn nha

bn ơi gõ laxte nha bn , khó hiểu quá!

Bài 1: 

a: Ta có: 5x=4y+2x

\(\Leftrightarrow3x=4y\)

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

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x+y}{4+3}=\dfrac{-56}{7}=-8\)

Do đó: x=-32; y=-24