dạng nhân này thì bạn đặt =k sẽ ra cả nha,mk ngại đánh lắm

40/x-30=20/y-15=28/z-21 => 40/x-40/30=20/y-20/15=28/z-28/21 => 40/x-4/3=20/y-4/3=28/z-4/3

<=> 40/x=20/y=28/z=K => x=40.K; y=20.K; z=28.K

<=> xyz=40.20.28.K³ => xyz=22400.K³ 

<=>K³=1 => K=+-1

<=> x=40.K = 40.1=40 (1)

                   =40.(-1)=-40

TH(1): x=40 => y=20; z =28

TH(2); x=-40 => y=-20; z=-28

vậy x=40; y=20; z =28

hoặc x=-40; y=-20; z=-28

câu b làm y vậy đó bạn đổi 15x=-10y=6z=>x/1/15=y/-1/10=z/1/6

1/2x X 1/3 x X + 1/6 x X = 28/21

ai giúp mk vs

Câu a và câu  b khó quá nên minh chí giúp bn câu b thôi!

c chứ ko phải b nha bn mình viết nhầm

\(\Rightarrow\left[\begin{array}{nghiempt}x-9=15k\\y-12=20k\\z-24=40k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=15k+9\\y=20k+12\\z=40k+24\end{array}\right.}\)

ta có:

x.y=1200\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\Rightarrow\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}=k\)

=> (15k+9)(20k+12)=1200

=> 3.4(5k+3)(5k+3)=1200

=> (5k+3)²=100

=> 5k+3=\(\pm\)10

 => \(\left[\begin{array}{nghiempt}5k+3=10\\5k+3=-10\end{cases}\Rightarrow\left[\begin{array}{nghiempt}5k=7\\5k=-13\end{cases}\Rightarrow}\left[\begin{array}{nghiempt}k=\frac{7}{5}\\k=-\frac{13}{5}\end{array}\right.}\)

* với k=7/5

x=7/5x15+9=30

y=7/5x20+12=40

z=7/5x40+24=80

* với k=-13/5

x=-13/5x15+9=-30

y=-13/5x20+12=-40

z=-13/5x40+24=-80

b)

\(\frac{40}{x-30}=\frac{20}{y-50}=\frac{28}{z-21}\Rightarrow\frac{x-30}{40}=\frac{y-50}{20}=\frac{z-21}{28}k=\)

=>\(\left[\begin{array}{nghiempt}x-30=40k\\y-50=20k\\z-21=28k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=40k+30\\y=20k+50\\z=28k+21\end{array}\right.}\)

ta có:

x.y.z=22400

=> (40k+30)(20k+50)(28k+21)=22400

c) 15x=-10y=6z

\(\Rightarrow\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\Rightarrow\frac{x}{2}=-\frac{y}{3}=\frac{z}{5}=k\)

=> \(\left[\begin{array}{nghiempt}x=2k\\y=-3k\\z=5k\end{array}\right.\)

ta có:

x.y.z=30000

=> 2k.(-3k).5k=30000

=> k³=1000

=> k=10

ta có: x=10x2=20

          y=10.(-3)=-30

          z=10.5=50

giúp mk vs ! mai đi hok ùi

mk ms lm đc câu a và c

\(a.\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\&xy=1200\)

\(\Leftrightarrow\dfrac{15}{20}=\dfrac{x-9}{y-12}\Leftrightarrow\dfrac{3}{4}=\dfrac{x-9}{y-12}\)

\(\Rightarrow\dfrac{9}{12}=\dfrac{x-9}{y-12}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{9}{12}=\dfrac{x-9}{y-12}=\dfrac{x-9+9}{y-12+12}=\dfrac{x}{y}\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{xy}{y^2}=\dfrac{x^2}{xy}\)

Từ \(\dfrac{3}{4}=\dfrac{xy}{y^{^2}}\Rightarrow\dfrac{3}{4}=\dfrac{1200}{y^2}\Rightarrow y^2=1200.\dfrac{4}{3}=1600\)

\(\Rightarrow y=\sqrt{1600}=\pm40\)

+ TH1: \(y=40\Rightarrow x=30\)

\(\dfrac{15}{x-9}=\dfrac{40}{z-24}\Rightarrow z=80\) (tự giải pt)

+ TH2: \(y=-40\Rightarrow x=-30\)

\(\dfrac{15}{x-9}=\dfrac{40}{z-4}\Rightarrow z=-80\) (tự giải pt)

Vậy, các cặp \(\left(x;y;z\right)\) thỏa mãn là \(\left(30;40;80\right)\&\left(-30;-40;-80\right)\)

\(b.15x=-10y=6z\&xyz=30000\)

\(\Rightarrow\left\{{}\begin{matrix}15x=-10y\\-10y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-10}=\dfrac{y}{15}\\\dfrac{y}{6}=\dfrac{z}{-10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-20}=\dfrac{y}{30}\\\dfrac{y}{30}=\dfrac{z}{-50}\end{matrix}\right.\Rightarrow\dfrac{x}{-20}=\dfrac{y}{30}=\dfrac{z}{-50}\)

Đặt \(\dfrac{x}{-20}=\dfrac{y}{30}=\dfrac{z}{-50}=k\Rightarrow x=-20k;y=30k;z=-50k\)

\(\Rightarrow xyz=30000\Rightarrow-20k.30k.\left(-50k\right)=30000\Rightarrow30000k^3=30000\)

\(\Rightarrow k^3=1\Rightarrow k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=30\\z=-50\end{matrix}\right.\)

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