1. -2x=5y =>\(\frac{x}{y}=\frac{-5}{2}=>y=\frac{-2x}{5}\)

Thế y=\(\frac{-2x}{5}\) ta được:

x+\(\frac{-2x}{5}\)=30     \(\Rightarrow\frac{5x-2x}{5}=30\)

\(\Rightarrow3x=150\)\(\Rightarrow x=50\)

=>y=30-x=30-50=-20.

Vậy x=50; y=-20.

Những bài khác tương tự bạn nhé!

bạn kia làm đúng rồi

k tui nha 

thank

1,7y

Từ hệ thứ 2: \(\left\{{}\begin{matrix}3x+5y=7\\2x-y=2m\end{matrix}\right.\)

So sánh với hệ thứ nhất, ta thấy 2 hệ tương đương khi và chỉ khi \(2m=6\)

\(\Leftrightarrow m=3\)

a: \(\Leftrightarrow-15x+10=-7x+14\)

=>-8x=4

hay x=-1/2

\(a,\dfrac{2-3x}{x-2}=-\dfrac{7}{5}\left(x\ne2\right)\\ \Leftrightarrow14-7x=10-15x\\ \Leftrightarrow8x=-4\Leftrightarrow x=-2\left(tm\right)\\ c,\Leftrightarrow\dfrac{x-1}{2}=\dfrac{y-2}{5}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{2\cdot2+5\cdot3-4}=\dfrac{45}{15}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=6\\y-2=15\\z-3=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=17\\z=15\end{matrix}\right.\\ d,\Leftrightarrow\dfrac{x}{1}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{x}{4}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{6x+7y+8z}{24+84+120}=\dfrac{456}{228}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x=8\\y=24\\z=30\end{matrix}\right.\)

Áp dụng t/c dtsbn:

\(\dfrac{x+3}{5}=\dfrac{y-2}{3}=\dfrac{z-1}{7}=\dfrac{3x+9}{15}=\dfrac{5y-10}{15}=\dfrac{7z-7}{49}=\dfrac{3x-5y+7z+9+10-7}{15-15+49}=\dfrac{86+12}{49}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x+3=2.5=10\\y-2=2.3=6\\z-1=2.7=14\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=10-3=7\\y=6+2=8\\z=14+1=15\end{matrix}\right.\)

cứ đặt k là giải đc

ket bn vs mk , mk giai ky cho ^_^

\(a,3x=5y\)và \(xy=60\)

     \(3x=5y\)

\(\Leftrightarrow\frac{x}{5}=\frac{y}{3}\)

\(\Leftrightarrow\frac{x^2}{25}=\frac{xy}{15}=\frac{y^2}{9}\)

\(\Leftrightarrow\frac{x^2}{25}=\frac{y^2}{9}=\frac{60}{15}=4\)

\(\Rightarrow\hept{\begin{cases}\frac{x^2}{25}=4\\\frac{y^2}{9}=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=100\\y^2=36\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\pm10\\y=\pm6\end{cases}}\)

  Vậy \(\left(x,y\right)\in\left\{\left(-10,-6\right);\left(10,6\right)\right\}\)

\(b,4x=5y\)và \(x^2-y^2=9\)

       \(4x=5y\)

\(\Leftrightarrow\frac{x}{5}=\frac{y}{4}\)

\(\Leftrightarrow\frac{x^2}{25}=\frac{y^2}{16}\)

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

  \(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{9}{9}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{x^2}{25}=1\\\frac{y^2}{16}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=25\\y^2=16\end{cases}\Leftrightarrow\hept{\begin{cases}x=\pm5\\y=\pm4\end{cases}}}\)

   Vậy \(\left(x,y\right)\in\left\{\left(-5,-4\right);\left(5,4\right)\right\}\)

 \(c,x:3=y:7\)và xy = 21

       \(x:3=y:7\)

\(\Leftrightarrow\frac{x}{3}=\frac{y}{7}\)

\(\Leftrightarrow\frac{x^2}{9}=\frac{xy}{21}=\frac{y^2}{49}\)

\(\Leftrightarrow\frac{x^2}{9}=\frac{y^2}{49}=\frac{21}{21}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{x^2}{9}=1\\\frac{y^2}{49}=1\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=9\\y^2=49\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\pm3\\y=\pm7\end{cases}}}\)

 Vậy \(\left(x,y\right)\in\left\{\left(-3,-7\right);\left(3,7\right)\right\}\)

\(d,2x=9y\)và xy = 72

      \(2x=9y\)

\(\Leftrightarrow\frac{x}{9}=\frac{y}{2}\)

\(\Leftrightarrow\frac{x^2}{81}=\frac{xy}{18}=\frac{y^2}{4}\)

\(\Leftrightarrow\frac{x^2}{81}=\frac{y^2}{4}=\frac{72}{18}=4\)

\(\Rightarrow\hept{\begin{cases}\frac{x^2}{81}=4\\\frac{y^2}{4}=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=324\\y^2=16\end{cases}\Leftrightarrow\hept{\begin{cases}x=\pm18\\y=\pm4\end{cases}}}\)

 Vậy \(\left(x,y\right)\in\left\{\left(-18,-4\right);\left(18,4\right)\right\}\)

đặt x+3/5=y-2/3=z-1/7=k

=> x=5k-3  ; y=3k+2   ;  z=7k+1

ta có:

3(5k-3)-5(3k+2)+7(7k+1)=86

15k-9-15k-10+49k+7=86

49k-12=86

49k=98

k=2

ta có: x=2x5-3=7

y=2x3+2=8

z=2x7+1=15

a)

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{3x-2y}{3.5-2.2}=\dfrac{-55}{11}=-5\)

=> \(\left\{{}\begin{matrix}x=-5.5=-25\\y=-5.2=-10\end{matrix}\right.\)

b)

\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{2x+5y}{2.3+5.2}=\dfrac{48}{16}=3\)

=> \(\left\{{}\begin{matrix}x=3.3=9\\y=3.2=6\end{matrix}\right.\)

c)

Có: \(\dfrac{x}{y}=-\dfrac{5}{2}\Leftrightarrow-\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x+y}{-5+2}=\dfrac{30}{-3}=-10\)

=> \(\left\{{}\begin{matrix}x=-10.-5=50\\y=-10.2=-20\end{matrix}\right.\)

d)

Có: \(\dfrac{x}{y}=\dfrac{4}{3}\Leftrightarrow\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{2x+3y}{2.4+3.3}=\dfrac{34}{17}=2\)

=> \(\left\{{}\begin{matrix}x=2.4=8\\y=2.3=6\end{matrix}\right.\)