Giải:

Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow\left\{\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)

Ta có: \(a^2+3b^2-2c^2=\left(-16\right)\)

\(\Rightarrow4k^2+27k^2-32k^2=-16\)

\(\Rightarrow\left(-1\right)k^2=-16\)

\(\Rightarrow k^2=16\)

\(\Rightarrow k=\pm4\)

+) \(k=4\Rightarrow a=8;b=12;c=16\)

+) \(k=-4\Rightarrow a=-8;b=-12;c=-16\)

Vậy bộ số \(\left(x;y;z\right)\) là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)

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 : 

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

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

Áp dụng BĐt cô-si, ta có \(\frac{2\left(a+b\right)^2}{2a+3b}\ge\frac{8ab}{2a+3b}=\frac{8}{\frac{2}{b}+\frac{3}{a}}\)

                                      \(\frac{\left(b+2c\right)^2}{2b+c}\ge\frac{8bc}{2b+c}=\frac{8}{\frac{2}{c}+\frac{1}{b}}\)

                                        \(\frac{\left(2c+a\right)^2}{c+2a}\ge\frac{8ac}{c+2a}\ge\frac{8}{\frac{1}{a}+\frac{2}{c}}\)

Cộng 3 cái vào, ta có 

A\(\ge8\left(\frac{1}{\frac{2}{b}+\frac{3}{a}}+\frac{1}{\frac{1}{b}+\frac{2}{c}}+\frac{1}{\frac{1}{a}+\frac{2}{c}}\right)\ge8\left(\frac{9}{\frac{3}{b}+\frac{4}{c}+\frac{4}{a}}\right)=8.\frac{9}{3}=24\)

Vậy A min = 24 

Neetkun ^^

bạn tìm ra dấu= xảy ra khi nào

rất tiếc sai rồi :)) 

Bạn có làm trong này rồi nhé Câu hỏi của Phạm Vũ Ngọc Duy

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\\ \Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\\ =\dfrac{a^2+3b^2-2c^2}{4+27-32}=-\dfrac{16}{-1}=16\\ \Rightarrow a=\pm8;b=\pm12;c=\pm16\)

Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

\(\Rightarrow\dfrac{a^2}{4}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}\)

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

\(\dfrac{a^2}{4}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=\dfrac{-16}{-1}=16\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{a^2}{4}=16\Rightarrow a=8\\\dfrac{3b^2}{27}=16\Rightarrow b=12\\\dfrac{2c^2}{32}=16\Rightarrow c=16\end{matrix}\right.\)

b) Ta có : \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\)

\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Khi đó \(a=12.\dfrac{3}{2}=18;b=12.\dfrac{4}{3}=16;c=12.\dfrac{5}{4}=15\)

Vậy (a,b,c) = (18,16,15) 

tham khảo!!

https://lazi.vn/edu/exercise/tim-cac-so-a-b-c-biet-rang-a-2-b-3-c-4-va-a-2-b-2-2c-2-108

Giải:

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)

Ta có: \(a^2+3b^2-2c^2=-16\)

\(\Rightarrow4k^2+27k^2-32k^2=-16\)

\(\Rightarrow-k^2=-16\)

\(\Rightarrow k^2=16\)

\(\Rightarrow k=\pm4\)

+) \(k=4\Rightarrow a=8,b=12,c=16\)

+) \(k=-4\Rightarrow a=-8;b=-12;c=-16\)

Vậy bộ số \(\left(a;b;c\right)\) là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)

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

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

\(\dfrac{a^2}{4}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=\dfrac{-16}{-1}=16\)

\(\Rightarrow a^2=64,b^2=144,c^2=256\) hay:

\(\left(a;b;c\right)=\left(8;12;16\right)=\left(-8;-12;-16\right)\)

ĐS: \(\left(a;b;c\right)=\left(8;12;16\right)=\left(-8;-12;-16\right)\)

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)

Ta có: \(a^2+3b^2-2c^2=-16\)

\(\Leftrightarrow\left(2k\right)^2+3\left(3k\right)^2-2\left(4k\right)^2=-16\)

\(\Leftrightarrow4k^2+3\cdot9k^2-2\cdot16k^2=-16\)

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

\(\Leftrightarrow\left(4+27-32\right)k^2=-16\)

\(\Leftrightarrow-k^2=-16\)

\(\Leftrightarrow k^2=16\)

\(\Leftrightarrow k=\sqrt{16}\)

\(\Leftrightarrow k=\pm4\)

Nếu \(k=-4\) thì:

\(a=2\left(-4\right)=-8\)

\(b=3\left(-4\right)=-12\)

\(c=4\left(-4\right)=-16\)

Nếu \(k=4\) thì :

\(a=2\cdot4=8\)

\(b=3\cdot4=12\)

\(c=4\cdot4=16\)

Vậy \(a;b;c=\left\{-8;-12;-16\right\}\) hoặc \(a;b;c=\left\{8;12;16\right\}\)