\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4.\)

Mình nghĩ bạn t hiếu cái gì đó thì phải

Ta có:\(\dfrac{x^2}{4}=\dfrac{x}{2};\dfrac{y^2}{9}=\dfrac{y}{3};\dfrac{z^2}{25}=\dfrac{z}{5}\)

Aps dụng tính chất dãy tỉ số bằn nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)

=>\(\dfrac{x}{2}=1=>x=2\)

  \(\dfrac{y}{3}=1=>y=3\)

\(\dfrac{z}{5}=1=>z=5\)

Vậy x=2, y=3, z=5

Ta có : \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)

\(\Leftrightarrow x=2;y=3;z=5\)

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)

=> a² = 16

=> a = 4 hoặc a = -4

Thay vào \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\) tìm nốt a, b, c

hjhj, thật ra bài này mik làm đc. mik gửi cho vui thôi

dù gì thì

a) \(\dfrac{a}{5}=\dfrac{b}{4}\Rightarrow\dfrac{a^2}{25}=\dfrac{b^2}{16}\)

Áp dụng tính chất DTSBN :

\(\dfrac{a^2}{25}=\dfrac{b^2}{16}=\dfrac{a^2-b^2}{25-16}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{1}{9}\cdot25=\dfrac{25}{9}\\b^2=\dfrac{1}{9}\cdot16=\dfrac{16}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{5}{3};b=\dfrac{4}{3}\\a=\dfrac{-5}{3};b=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(\left(a;b\right)\in\left\{\left(\dfrac{5}{3};\dfrac{4}{3}\right);\left(-\dfrac{5}{3};-\dfrac{4}{3}\right)\right\}\)

b) \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)

Áp dụng tính chất DTSBN :

\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=4.4=16\\b^2=4.9=36\\c^2=4,16=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=4;=6;c=8\\a=-4;b=-6;c=-8\end{matrix}\right.\)

Vậy (a;b;c) \(\in\left\{\left(4;6;8\right);\left(-4;-6;-8\right)\right\}\)

1. \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}a=5\times2=10\\b=5\times3=15\\c=5\times4=20\end{matrix}\right.\)

1. \(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\Rightarrow a=10;b=15;c=20\)

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

\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a-b-c}{8-12-15}=\dfrac{28}{-19}=\dfrac{-28}{19}\)

Do đó: \(\left\{{}\begin{matrix}a=\dfrac{-224}{19}\\b=\dfrac{-336}{19}\\c=\dfrac{-420}{19}\end{matrix}\right.\)

Đặt a/2=b/3=c/4=k

=>a=2k; b=3k; c=4k

Ta có: \(a^2+3b^2-2c^2=-16\)

\(\Leftrightarrow4k^2+27k^2-32k^2=-16\)

\(\Leftrightarrow k^2=16\)

Trường hợp 1: k=4

=>a=8; b=12; c=16

Trường hợp 2: k=-4

=>a=-8; b=-12; c=-16

cảm ơn

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+2b-3c}{2+2\cdot3-3\cdot4}=\dfrac{-20}{-4}=5\\ \Rightarrow\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=5\\\dfrac{b}{3}=5\\\dfrac{c}{4}=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)