Bài 2:

\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)

\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)

a) Ta có: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)

nên \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

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

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{2x}{3}=12\\\dfrac{3y}{4}=12\\\dfrac{4z}{5}=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=36\\3y=48\\4z=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=20\end{matrix}\right.\)

Vậy: (x,y,z)=(18;16;20)

b) Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\)

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

Ta có: \(x^2-y^2=4\)

\(\Leftrightarrow\left(5k\right)^2-\left(3k\right)^2=4\)

\(\Leftrightarrow16k^2=4\)

\(\Leftrightarrow k\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

Trường hợp 1: \(k=\dfrac{1}{2}\)

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

Trường hợp 2: \(k=-\dfrac{1}{2}\)

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

Vậy: \(\left(x,y\right)\in\left\{\left(\dfrac{5}{2};\dfrac{3}{2}\right);\left(-\dfrac{5}{2};-\dfrac{3}{2}\right)\right\}\)

a)

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

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Suy ra : 

\(x=\dfrac{12.3}{2}=18\\ y=\dfrac{12.4}{3}=16\\ z=\dfrac{12.5}{4}=15\)

b)

\(x=\dfrac{y}{3}.5=\dfrac{5y}{3}\\ x^2-y^2=4\\ \Leftrightarrow\left(\dfrac{5y}{3}\right)^2-y^2=4\\ \Leftrightarrow\dfrac{16y^2}{9}=4\Leftrightarrow y=\pm\dfrac{3}{2} \)

Với $y = \dfrac{3}{2}$ thì $x = \dfrac{5}{2}$

Với $y = \dfrac{-3}{2}$ thì $x = \dfrac{-5}{2}$

c)

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

Suy ra : 

\(2x=y+z+1\Leftrightarrow y+z=2x-1\)

Mặt khác : 

\(x+y+z=\dfrac{1}{2}\Leftrightarrow x+2x-1=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)

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

Mà \(y+z=0\Leftrightarrow z=-y\)

nên suy ra:  \(y=\dfrac{1}{2};z=-\dfrac{1}{2}\)

1. Áp dụng TCDTSBN ta có:

$\frac{x-1}{3}=\frac{y-2}{4}=\frac{z+5}{6}=\frac{x-1+(y-2)-(z+5)}{3+4-6}$

$=\frac{x+y-z-8}{1}=\frac{8-8}{1}=0$

$\Rightarrow x-1=y-2=z+5=0$

$\Rightarrow x=1; y=2; z=-5$

2.

Có:

$\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{6}=\frac{2x+2}{4}=\frac{3y+9}{12}=\frac{4z+20}{24}$

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

$\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{6}=\frac{2x+2}{4}=\frac{3y+9}{12}=\frac{4z+20}{24}=\frac{2x+2+3y+9+4z+20}{4+12+24}=\frac{2x+3y+4z+31}{40}=\frac{9+31}{40}=1$

Suy ra:

$x+1=2.1=2\Rightarrow x=1$

$y+3=1.4=4\Rightarrow y=1$

$z+5=6.1=6\Rightarrow z=1$

$

3.

Có:

$\frac{x+1}{3}=\frac{y+2}{-4}=\frac{z-3}{5}=\frac{3x+3}{9}=\frac{2y+4}{-8}=\frac{4z-12}{20}$

Áp dụng TCDTSBN:

$\frac{x+1}{3}=\frac{y+2}{-4}=\frac{z-3}{5}=\frac{3x+3}{9}=\frac{2y+4}{-8}=\frac{4z-12}{20}=\frac{3x+3+2y+4+4z-12}{9+(-8)+20}=\frac{3x+2y+4z-5}{21}=\frac{47-5}{21}=2$

Suy ra:

$x+1=3.2=6\Rightarrow x=5$

$y+2=(-4).2=-8\Rightarrow y=-10$

$z-3=5.2=10\Rightarrow z=13$

\(\frac{y+1}{4x^2+1}=1-\frac{4x^2-y}{4x^2+1}\ge1-\frac{4x^2-y}{2\sqrt{4x^2.1}}=1+\frac{y}{4x}-x;\)

Tương tự ta được \(\frac{1+z}{4y^2+1}\ge1+\frac{z}{4y}-y\); \(\frac{1+x}{4z^2+1}\ge1+\frac{x}{4z}-z\)

cộng 3 bất đăng thức trên ta được p \(\ge3+\frac{1}{4}\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)-\left(x+y+z\right)=\frac{3}{2}+\frac{1}{4}\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)\ge\)\(\frac{3}{2}+\frac{1}{4}.3\sqrt[3]{\frac{y}{x}.\frac{z}{y}.\frac{x}{z}}=\frac{9}{4}\)

p min khi x=y=z = 1/2

 x,y,z khác 0 
Từ 2x=-3y => x=-3/2y và -3y =4z => z=-3/4y thay vào pt đầu ta được 
-2/3y + 1/y - 4/3y =3 <=> y=-1/3 => x=1/2 và z =1/4