1/ Ta có xy=-6

Với x=-6 => y=1

x=-3 => y=2 

x= -2 => y=3

x=-1 => y=6

2/ Ta có x=y+4 

Thay x=y+4 vào bt, ta được 

<=> y+4-3/y-2 =3/2

<=> y+1/y-2=3/2

<=> 2(y+1)=3(y-2)

<=> 2y +2 = 3y - 6

<=> 3y - 2y= 2+ 6

<=> y= 8 <=> x= 12

3/ -4/8 = x/-10 <=> x= (-4)*(-10)/8=5

-4/8 = -7/y <=> y=(-7)*8/(-4) =14

-4/8 = z/-24 <=> z= (-4)*(-24)/8=12

a)\(\dfrac{x}{60}=-\dfrac{3}{4}\)

\(\Rightarrow x\cdot4=60\cdot\left(-3\right)\)

   \(x\cdot4=-180\)

  x=45

b)\(\dfrac{2}{5}=\dfrac{12}{x}\)

   \(\Rightarrow2x=5\cdot12\)

      \(2x=60\)

       x=30

c)\(x-\dfrac{5}{7}=\dfrac{6}{21}\)

      \(x=\dfrac{2}{7}+\dfrac{5}{7}\)

      x=1

d)\(x+\dfrac{7}{8}=\dfrac{63}{24}\)

     \(x=\dfrac{21}{8}-\dfrac{7}{8}\)

     \(\dfrac{14}{8}\)

a)\(\dfrac{x}{60}=\dfrac{-3}{4}\Rightarrow x=\dfrac{-3.60}{4}=-45\)

b)\(\dfrac{2}{5}=\dfrac{12}{x}\Rightarrow x=\dfrac{5.12}{2}=30\)

ngu như con lợn lò

5x7x77-7x60+49x25-15x42=2870

thằng hâm

=27195 - 420 + 1225 - 630

=27370

\(a.60\times\left(\frac{7}{12}+\frac{4}{15}\right)\)

\(=60\times\left(\frac{35}{60}+\frac{16}{60}\right)\)

\(=60\times\frac{51}{60}\)

\(=51\)

\(b.\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}\times\frac{8}{9}\)

\(=\frac{1\times2\times3\times4\times5\times6\times7\times8}{2\times3\times4\times5\times6\times7\times8\times9}\)

\(=\frac{1}{9}\)

-2x - 40 = (5 - x) - (-15 + 60)

-2x - 40 = (5 - x) - 45

-2x - 40 = -40 - x

-2x - x = 40 - 40

      -3x = 0

         x = 0

2(x - 4) - 3(x + 7) = 14

    2x - 8 - 3x - 21 = 14

                 -x - 29 = 14

                        -x = 14  + 29

                        -x = 43 

                         x = -43

-7(5 - x) - 2(x - 10) = 15

-35 + 7x - 2x + 20 = 15

              -15 + 5x = 15

                       5x = 15 + 15

                       5x = 30

                         x = 6

         \(-2x-40=\left(5-x\right)-\left(-15+60\right)\)

\(\Leftrightarrow\)\(-2x-40=5-x+15-60\)

\(\Leftrightarrow\)\(-2x-40=-x-40\)

\(\Leftrightarrow\)\(-2x+x=40-40\)

\(\Leftrightarrow\)\(x=0\)

Vậy....

\(x+18=2-2x\)

\(2x+x=2-18\)

\(3x=-20\)

\(x=-20:3\)

\(x=\frac{-20}{3}\)

Đề sai rồi bạn

a)\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{5}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{z^2}{25}\)

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

\(\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{z^2}{25}=\dfrac{x^2-y^2+z^2}{9-49+25}=\dfrac{-60}{-15}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.9=36\\y^2=4.49=196\\z^2=4.25=100\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm14\\z=\pm10\end{matrix}\right.\)

b)

\(\left\{{}\begin{matrix}5x=2y\\3y=5z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{5}\\\dfrac{y}{5}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{3}\)

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

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y+z}{2+5+3}=\dfrac{-60}{10}=-6\)

\(\Rightarrow\left\{{}\begin{matrix}x=-6.2=-12\\y=-6.5=-30\\z=-6.3=-18\end{matrix}\right.\)

a. \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{5}\) và \(x^2-y^2+z^2=-60\)

Theo t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{5}=\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{z^2}{25}=\dfrac{x^2-y^2+z^2}{9-49+25}=\dfrac{-60}{-15}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=4\Rightarrow x=4.3=12\\\dfrac{y}{7}=4\Rightarrow y=4.7=28\\\dfrac{z}{5}=4\Rightarrow z=5.4=20\end{matrix}\right.\)

Vậy...........

b. Đề bài câu b chắc là sai . mk sửa lại

\(5x=2y;3y=5z\) và \(x+y+z=-60\)

Theo đề bài ta có :

\(\begin{matrix}5x=2y\\3y=5z\end{matrix}\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{5}\\\dfrac{y}{5}=\dfrac{z}{3}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{3}\)

Áp dụng t/c dãy tỉ số bằng nhau :

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y+z}{2+5+3}=\dfrac{-60}{10}=-6\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-6\Rightarrow x=2.\left(-6\right)=-12\\\dfrac{y}{5}=-6\Rightarrow y=5.\left(-6\right)=-30\\\dfrac{z}{3}=-6\Rightarrow z=3.\left(-6\right)=-18\end{matrix}\right.\)

Vậy..........