Đặt \(\frac{x}{4}=\frac{y}{9}=k\Rightarrow x=4k;y=9k\)

\(\Rightarrow xy=144\Leftrightarrow4k\cdot9k=144\)

\(\Rightarrow36k^2=144\)

\(\Rightarrow k^2=4\Rightarrow k=\pm2\)

Nếu \(k=2\Rightarrow\hept{\begin{cases}x=4k=4\cdot2=8\\y=9k=9\cdot2=18\end{cases}}\)

Nếu \(k=-2\Rightarrow\hept{\begin{cases}x=4k=4\cdot\left(-2\right)=-8\\y=9k=9\cdot\left(-2\right)=-18\end{cases}}\)

mk hoc lop 6 soryy ko giup duoc

a,\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\Leftrightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)=3  

Bài 1: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12};\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\)

=>\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2z}{18}=\frac{3y}{36}\)

Áp dụng tính chất của dãy tỉ số bằng nhau: \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2z}{18}=\frac{3y}{36}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

=>x=27;z=36;z=60

Bài 2: \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=5k\end{cases}}\Rightarrow xy=2k.5k=10k^2=40\Rightarrow k^2=4\Rightarrow\hept{\begin{cases}k=-2\\k=2\end{cases}}\)

+)k=-2 => x=-4;y=-5

+)k=2 => x=4;y=5

Vậy x=-4;y=-5 hoặc x=4;y=5

Đặt  \(\frac{x}{4}=\frac{y}{5}=k\left(k\ne0\right)\)

\(\Rightarrow\hept{\begin{cases}x=4k\\y=5k\end{cases}}\)

Mà  \(xy=20\)\(\Leftrightarrow4k.5k=20\)

                            \(\Leftrightarrow20k^2=20\)

                            \(\Leftrightarrow k^2=1\)

                            \(\Leftrightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)

+) Với  \(k=1\Leftrightarrow\hept{\begin{cases}x=4\\y=5\end{cases}}\)

+) Với  \(k=-1\Leftrightarrow\hept{\begin{cases}x=-4\\y=-5\end{cases}}\)

Vậy ...

\(\frac{x}{4}=\frac{y}{5}\Rightarrow\frac{x^2}{4^2}=\frac{x}{4}.\frac{x}{4}=\frac{x}{4}.\frac{y}{5}=\frac{xy}{20}=\frac{20}{20}=1\)

\(\Rightarrow x^2=4^2=16\Rightarrow x=\pm4\)

Với x=4 thì \(y=\frac{4}{4}.5=5\)

Với x=-4 thì \(y=\frac{-4}{4}.5=-5\)

Có \(\frac{x}{4}=\frac{y}{5}\left(x.y=20\right)\)

\(\Rightarrow\hept{\begin{cases}x=4k\\y=5k\end{cases}}\left(k=0\right)\)

\(\Rightarrow4k.5k=x.y=20\)

\(\Rightarrow20k^2=20\)

\(\Rightarrow k^2=1\)

\(\Rightarrow k\in\left\{1;-1\right\}\)

+ TH1:

Nếu k = 1 thì:

x = 4

y = 5

+ TH2:

Nếu k = -1 thì:

x = -4

y = -5

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

Đặt: \(\frac{x}{4}=\frac{y}{7}=k\)

\(\Rightarrow x=4k;y=7k\)

\(\Rightarrow xy=4k.7k=112\)

\(\Rightarrow28k^2=112\)

\(\Rightarrow k^2=\frac{112}{28}=4\)

\(\Rightarrow\left[\begin{array}{nghiempt}k=2\\k=-2\end{array}\right.\)

Với \(k=2\) \(\Rightarrow\left[\begin{array}{nghiempt}x=4k=2.4=8\\y=7k=2.7=14\end{array}\right.\)

Với \(k=-2\Rightarrow\left[\begin{array}{nghiempt}x=4k=-2.4=-8\\y=7k=-2.7=-14\end{array}\right.\)

Đặt \(\frac{x}{4}=\frac{y}{7}\) = k

=> x = 4k; y = 7k

Ta thay vào: x . y = 112

=> 4k . 7k = 112

=> 28 . k² = 112

=> k² = 112 : 28

=> k² = 4

=> k = 2 hoặc k  = -2

Nếu k = 2 => x = 4 . 2 = 8; y = 7k = 7 . 2 = 14

Nếu k = -2 => x = 4 . (-2) = -8; y = 7 . (-2) = -14

Vậy x = {-8; 8} và y = {-14; 14}

Đặt \(\frac{x}{4}=\frac{y}{7}=k\\ x=4k;y=7k\)

Thay x.y = 112

=> 4k . 7k = 112

     28k²    = 112

         k²    = 112 : 28

         k²    = 4

=> k = 2 hoặc k = -2

Với k = 2 Suy ra x = 4.2= 8 ; y= 7.2= 14

Với k = (-2) Suy ra x= 4.(-2)= -8 ; y= 7.(-2)= -14
 

\(\frac{15}{x-9}=\frac{20}{y-12}\Rightarrow\frac{x-9}{15}=\frac{y-12}{20}\Leftrightarrow\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}\)

\(\Rightarrow\frac{x^2}{15^2}=\frac{x}{15}.\frac{y}{20}=\frac{1200}{300}=4=2^2\Rightarrow x^2=2^2.15^2=30^2\)

\(\Rightarrow x=30\text{ hoặc }x=-30\)

+TH1: x = 30

\(\frac{y}{20}=\frac{x}{15}\Rightarrow y=\frac{20.x}{15}=\frac{20.30}{15}=40\)

\(\frac{40}{z-24}=\frac{15}{30-9}=\frac{5}{7}\Rightarrow z=\frac{40.7}{5}+24=80\)

+TH2: x = -30

\(\frac{y}{20}=\frac{x}{15}=-\frac{30}{15}=-2\Rightarrow y=-2.20=-40\)

\(\frac{40}{z-24}=\frac{15}{-30-9}=-\frac{15}{3}\Rightarrow z=\frac{-3.40}{15}+24=16\)