a,\(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\Leftrightarrow\frac{3x}{12}=\frac{2y}{4}=\frac{4z}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{3x}{12}=\frac{2y}{4}=\frac{4z}{12}=\frac{3x-2y+4z}{12-4+12}=\frac{20}{20}=1\)

Suy ra:\(\hept{\begin{cases}\frac{x}{4}=1\\\frac{y}{2}=1\\\frac{z}{3}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=2\\z=3\end{cases}}\)

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

\(\frac{x}{2}=\frac{y}{6}=\frac{x-y}{2-6}=\frac{10}{-4}=-\frac{5}{2}\)

Suy ra:\(\hept{\begin{cases}\frac{x}{2}=-\frac{5}{2}\\\frac{y}{6}=-\frac{5}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-15\end{cases}}}\)

a) Đặt\(\frac{x}{5}=\frac{y}{2}=\frac{z}{3}=k.\)

Ta có : x = 5k ;  y = 2k ; z = 3k và xyz = 240

=> 5k . 2k . 3k = 240

=> k³ . 30 = 240

=> k³ = 8

=> k = 2

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=2\Leftrightarrow x=10\\\frac{y}{2}=2\Leftrightarrow y=4\\\frac{z}{3}=2\Leftrightarrow x=6\end{cases}}\)  

Vậy : x = 10; y = 4; z = 6

b) \(\frac{x}{4}=\frac{y}{3}=\frac{z}{2}\Rightarrow\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{4}\)

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

\(\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{4}=\frac{x^2-y^2-z^2}{16-9-4}=\frac{12}{3}=4\)

Suy ra :

\(\frac{x^2}{16}=4\Leftrightarrow x^2=64\Leftrightarrow x=\pm8\)

\(\frac{y^2}{9}=4\Leftrightarrow y^2=36\Leftrightarrow y=\pm6\)

\(\frac{z^2}{4}=4\Leftrightarrow z^2=16\Leftrightarrow z=\pm4\)

Vậy \(\hept{\begin{cases}x=8\\y=6\\z=4\end{cases}}\)hoặc \(\hept{\begin{cases}x=-8\\y=-6\\z=-4\end{cases}}\)

c) \(\frac{x}{4}=\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}\)

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

\(\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{x^2+y^2+z^2}{16+9+25}=\frac{200}{50}=4\) 

Suy ra :

\(\frac{x^2}{16}=4\Leftrightarrow x^2=64\Leftrightarrow x=\pm8\)

\(\frac{y^2}{9}=4\Leftrightarrow y^2=36\Leftrightarrow y=\pm6\)

\(\frac{z^2}{25}=4\Leftrightarrow z^2=100\Leftrightarrow z=\pm10\)

Vậy :\(\hept{\begin{cases}x=8\\y=6\\z=10\end{cases}}\)hoặc \(\hept{\begin{cases}x=-8\\y=-6\\z=-10\end{cases}}\)

a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)

b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)

\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)

\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)

a) $\genfrac{}{}{0ex}{}{x}{2}=\genfrac{}{}{0ex}{}{y}{5}=\genfrac{}{}{0ex}{}{z}{7};x+y+z=56$

$\genfrac{}{}{0ex}{}{x}{2}=\genfrac{}{}{0ex}{}{y}{5}=\genfrac{}{}{0ex}{}{z}{7}=\genfrac{}{}{0ex}{}{x+y+z}{2+5+7}=\genfrac{}{}{0ex}{}{56}{14}=4$

$\Rightarrow \{\begin{array}{c}x=4.2=8\\ y=4.5=20\\ z=4.7=28\end{array}$

b) $\genfrac{}{}{0ex}{}{x}{1,1}=\genfrac{}{}{0ex}{}{y}{1,3}=\genfrac{}{}{0ex}{}{z}{1,4}\left(1\right);2x-y=5,5$

$\left(1\right)\Rightarrow \genfrac{}{}{0ex}{}{2x-y}{1,1.2-1,3}=\genfrac{}{}{0ex}{}{5,5}{0,9}$

$\Rightarrow \{\begin{array}{c}x=1,1.\genfrac{}{}{0ex}{}{5,5}{0,9}=\genfrac{}{}{0ex}{}{6,05}{0,9}\\ y=1,3.\genfrac{}{}{0ex}{}{5,5}{0,9}=\genfrac{}{}{0ex}{}{7,15}{0,9}\\ z=\genfrac{}{}{0ex}{}{1,4}{1,1}.x=\genfrac{}{}{0ex}{}{1,4}{1,1}.\genfrac{}{}{0ex}{}{6,05}{0,9}=\genfrac{}{}{0ex}{}{8,47}{0,99}\end{array}$

d) $\genfrac{}{}{0ex}{}{x}{2}=\genfrac{}{}{0ex}{}{x}{3}=\genfrac{}{}{0ex}{}{z}{5};xyz=-30$

$\genfrac{}{}{0ex}{}{x}{2}=\genfrac{}{}{0ex}{}{x}{3}=\genfrac{}{}{0ex}{}{z}{5}=\genfrac{}{}{0ex}{}{xyz}{\mathrm{2.3.5}}=\genfrac{}{}{0ex}{}{-30}{30}=-1$

$\Rightarrow \{\begin{array}{c}x=2.\left(-1\right)=-2\\ y=3.\left(-1\right)=-3\\ z=5.\left(-1\right)=-5\end{array}$
 

a)\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{21}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

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

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{98}{48}=\frac{49}{23}\)

suy ra :

\(\frac{x}{10}=\frac{49}{23}\Rightarrow x=\frac{490}{23}\)

\(\frac{y}{15}=\frac{49}{23}\Rightarrow y=\frac{735}{23}\)

\(\frac{z}{21}=\frac{49}{23}\Rightarrow z=\frac{1029}{23}\)

bạn xem lại đề ra số hơi xấu

câu b:

\(\dfrac{x}{2}=\dfrac{y}{6}\\ \dfrac{x}{2}=\dfrac{y}{6}=\dfrac{x-y}{2-6}=\dfrac{10}{-4}=\dfrac{5}{-2}\\ x=\dfrac{5}{-2}.2=\dfrac{10}{-2}=-5\\ y=\dfrac{5}{-2}.6=-15\)

câu a:

\(\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\\ \dfrac{3x}{3.4}=\dfrac{2y}{2.2}=\dfrac{4z}{4.3}\\ \)

ta có

\(\dfrac{3x}{3.4}=\dfrac{2y}{2.2}=\dfrac{4z}{4.3}=\dfrac{3x+2y+4z}{12+4+12}=\dfrac{20}{28}=\dfrac{5}{7}\\ x=\dfrac{5}{7}:4=\dfrac{5}{28}\\ y=\dfrac{5}{7}:2=\dfrac{5}{14}\\ z=\dfrac{5}{7}:3=\dfrac{5}{21}\)