1. Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)

+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)

+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)

+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)

Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)

2,Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)

+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)

+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)

+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)

Vậy \(x=-2;y=-3;c=-4\)

3,Từ \(5x=8y=20z\Rightarrow\dfrac{5x}{160}=\dfrac{8y}{160}=\dfrac{20z}{160}\)

\(\Rightarrow\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}\)

Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}=\dfrac{x-y-z}{32-20-8}=\dfrac{3}{4}\)

+\(\dfrac{x}{32}=\dfrac{3}{4}\Rightarrow x=\dfrac{32.3}{4}=24\)

+\(\dfrac{y}{20}=\dfrac{3}{4}\Rightarrow y=\dfrac{20.3}{4}=15\)

+\(\dfrac{z}{8}=\dfrac{3}{4}\Rightarrow z=\dfrac{3.8}{4}=6\)

Vậy \(x=24;y=15;z=6\)

bình phương cả 2 vế ta được

\(A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2x^2+2y^2+2z^2\)

\(A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\) (vì x^2 +y^2 +z^2 =1)

Áp dụng BĐT cô si cho 2 số

\(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\left(1\right)\)

\(\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\left(2\right)\)

\(\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\left(3\right)\)

(1)+(2)+(3)

=> \(2\left(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\right)\ge2\left(x^2+y^2+z^2\right)\)

<=> \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge1\)

Cộng 2 vào cả 2 vế ta đc

\(A^2\ge3\)

<=> \(\ge\sqrt{3}\)

Vậy Min A= \(\sqrt{3}\) khi x=y=z =\(\dfrac{1}{\sqrt{3}}\)

Lời giải khác:

Đặt \((\frac{xy}{z}; \frac{yz}{x}; \frac{xz}{y})\mapsto (a,b,c)\)

\(\Rightarrow (x^2,y^2,z^2)=(ac,ab,bc)\)

Bài toán trở thành tìm min của $A=a+b+c$ biết $ab+bc+ac=1$ và $a,b,c>0$

Theo hệ quả quen thuộc của BĐT AM-GM:
\(A^2=(a+b+c)^2\geq 3(ab+bc+ac)=3\)

\(\Rightarrow A\geq \sqrt{3}\)

Vậy \(A_{\min}=\sqrt{3}\Leftrightarrow a=b=c\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)

a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).

Áp dụng t/c dãy tỉ số = nhau, ta có :

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)

⇒2x = 3.30 = 90 ⇒ x = 45

3y = 3.60 = 180 ⇒ y = 60

z = 3.28 = 84

Ý b) có gì đó sai sai ?

c)Ta có :

\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Áp dụng t/c dãy tỉ số = nhau, ta có :

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

⇒x = 5.15 = 75

y = 5.10 = 50

z = 5.6 = 30

d)Ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)

⇒ x = 2k ; y = 3k ; z = 5k

⇒ xyz = 2k.3k.5k = 30k³ = 810

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{1}{2xy}\ge\dfrac{4}{1^2}+\dfrac{1}{\dfrac{2.\left(x+y\right)^2}{4}}\ge4+2=6\)

Dấu "=" xảy ra <=> x = y = 0,5

\(\dfrac{x}{5}=\dfrac{y}{3}\Leftrightarrow\dfrac{x^2}{25}=\dfrac{y^2}{9}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x^2}{25}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{25-9}=\dfrac{4}{16}=\dfrac{1}{4}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2}{25}=\dfrac{1}{4}\\\dfrac{y^2}{9}=\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2,5\\x=-2,5\end{matrix}\right.\\\left[{}\begin{matrix}y=1,5\\y=-1,5\end{matrix}\right.\end{matrix}\right.\)

Vậy ..

a)Ta có :

\(\dfrac{x}{2}=\dfrac{y}{5}=k\)

Mà x.y=3,6 => 2k+5k=3,6=>7k=3,6

Vậy k = \(\dfrac{18}{35}\)

\(x=2k\Rightarrow x=\dfrac{36}{35}\)

\(y=5k\Rightarrow y=\dfrac{18}{7}\)

\(a,\dfrac{x}{2}=\dfrac{y}{5}\)

\(\rightarrow\)\(x.5=y.2\)

\(x.x.5=y.x.2\)

\(x^2.5=3,6.2\)

\(x^2.5=7,2\)

\(x^2=1,44\)

\(\rightarrow x=1,2\) hoặc \(x=-1,2\)

Ý b bạn làm tường tự nha

Từ \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\)

Và \(\dfrac{y}{6}=\dfrac{z}{8}\Rightarrow\)\(\dfrac{y}{12}=\dfrac{z}{16}\)

Suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)\(\Rightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\)

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

\(\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=-1\Rightarrow x=-1\cdot9=-9\\\dfrac{y}{12}=-1\Rightarrow y=-1\cdot12=-12\\\dfrac{z}{16}=-1\Rightarrow z=-1\cdot16=-16\end{matrix}\right.\)

Ta có :

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x}{9}=\dfrac{y}{12}\)(1)

\(\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{y}{12}=\dfrac{z}{16}\)(2)

Từ (1) và (2) , suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)

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

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)

Do đó :

\(\dfrac{x}{9}=-1\Rightarrow x=-1.9=-9\)

\(\dfrac{y}{12}=-1\Rightarrow y=-1.12=-12\)

\(\dfrac{z}{16}=-1\Rightarrow z=-1.16=-16\)

Vậy x = -9 ; y = -12 ; z = -16

\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{6}=\dfrac{z}{8}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{16}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)

\(\Rightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\)

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

\(\dfrac{3x}{54}=\dfrac{y}{24}=\dfrac{z}{32}\)

\(=\dfrac{3x-2y-z}{27-24-16}\)

\(=\dfrac{13}{-13}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=9.-1=-9\\y=12.-1=-12\\z=16.-1=-16\end{matrix}\right.\)

Đặt :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

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

Thay \(\left(1\right)\) vào \(xyz=810\) ta dduocj :

\(2k.3k.5k=810\)

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

\(\Leftrightarrow k=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)

Vậy ..

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

mà xyz = 810

hay \(2k.3k.5k=810\)

\(\Rightarrow30.k^2=810\)

\(\Rightarrow k^2=27=3^3\)

\(\Rightarrow k=3\)

Với k = 3 \(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=3.3=9\\z=5.3=15\end{matrix}\right.\)

Vậy.........

a. Đặt \(\dfrac{x}{-3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=-3k\\y=5k\end{matrix}\right.\)

mà \(x.y=\dfrac{-5}{27}\)

hay \(-3k.5k=\dfrac{-5}{27}\)

\(\Rightarrow-15.k^2=\dfrac{-5}{27}\)

\(\Rightarrow k^2=\dfrac{1}{81}=\left(\pm\dfrac{1}{9}\right)^2\)

Với \(k=\dfrac{1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{5}{9}\end{matrix}\right.\)

Với \(k=\dfrac{-1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{-5}{9}\end{matrix}\right.\)

Vậy.......

b. Từ \(\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{3}=\dfrac{z}{5}\end{matrix}\) \(\Rightarrow\begin{matrix}\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\) \(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{x-y+z}{9-12+20}=\dfrac{32}{17}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{32}{17}\Rightarrow x=\dfrac{32.9}{17}=\dfrac{288}{17}\\\dfrac{y}{12}=\dfrac{32}{17}\Rightarrow y=\dfrac{32.12}{17}=\dfrac{384}{17}\\\dfrac{z}{20}=\dfrac{32}{17}\Rightarrow z=\dfrac{32.20}{17}=\dfrac{640}{17}\end{matrix}\right.\)

Vậy.........