Giải thích các bước giải:

Bài 3:

a) `x/y = (-5)/7 <=> x/(-5) = y/7`

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

`x/(-5) = y/7 = (x-y)/(-5-7) = (-84)/(-12) = 7`

$\Rightarrow \begin{cases} \dfrac{x}{-5}=7 \to x = -35\\ \dfrac{y}{7}=7 \to y = 49\end{cases}$

Vậy `x = -35; y = 49`

b) `3x = -4y <=> x/(-4) = y/3`

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

`x/(-4) = y/3 = (2x - y)/(2.(-4) - 3) = 33/(-11) = -3`

$\Rightarrow \begin{cases} \dfrac{x}{-4}=-3 \to x = 12\\ \dfrac{y}{3}=-3 \to y = -9\end{cases}$

Vậy `x=12;y=-9`

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

`x/2 = y/5 = z/4 = (2x - 3y + z)/(2.2 - 3.5 + 4) = 45/(-7)`

$\Rightarrow \begin{cases} \dfrac{x}{2}=\dfrac{45}{-7} \to x = \dfrac{-90}{7}\\ \dfrac{y}{5}=\dfrac{45}{-7} \to y = \dfrac{-225}{7}\\ \dfrac{z}{4}=\dfrac{45}{-7} \to z=\dfrac{-180}{7}\end{cases}$

Vậy `x= (-90)/7; y = (-225)/7; (-180)/7`

d) `x : y : z= 3:5:(-2) <=> x/3 = y/5 = z/(-2)`

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

`x/3 = y/5 = z/(-2) = (5x-y+3z)/(5 . 3 - 5 + 3 . (-2)) = 124/4 =31`

$\Rightarrow \begin{cases}\dfrac{x}{3}=31 \to x = 93\\ \dfrac{y}{5}=31 \to y =155\\\dfrac{z}{-2}=31 \to z = -62\end{cases}$

Vậy `x=93; y=155; z = -62`

e) $\begin{cases} \dfrac{x}{2}=\dfrac{y}{3}\to \dfrac{x}{10}=\dfrac{y}{15}\\\dfrac{y}{5}=\dfrac{z}{4}\to \dfrac{y}{15}=\dfrac{z}{12}\end{cases}$

`=>x/10=y/15=z/12`

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

`x/10=y/15=z/12 = (2x - 3y + 4z)/(2.10 - 3.15 + 4.12)=69/23 = 3`

$\Rightarrow \begin{cases}\dfrac{x}{10}=3 \to x=30\\\dfrac{y}{15}=3 \to y = 45\\\dfrac{z}{12}=3 \to z = 36\end{cases}$

Vậy `x=30;y=45;z=36`

g) `x/2 = y/3 = z/4 => x^2/4 = (3y^2)/27 = z^2/16`

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

`x^2/4 = (3y^2)/27 = z^2/16 = (x^2 + 3y^2 - z^2)/(4 + 27-16)= 60/15 = 4`

$\Rightarrow \begin{cases} \dfrac{x^2}{4} = 4 \to x^2 = 16 \to x = \pm4\\\dfrac{3y^2}{27}=4 \to 3y^2=108 \to y^2=36 \to y = \pm 6\\\dfrac{z^2}{16}=4\to z = 64 \to z = \pm 8\end{cases}$

Vậy........

f) `(x-1)/2 = (3-y)/3 = (5+z)/4 =  (2x-2)/4`  

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

`(x-1)/2 = (3-y)/3 = (5+z)/4 =  (2x-2)/4 = (2x - 2 - 3 + y -5-z)/(4 - 3 - 4) = ((2x + y -z) -10)/(-3) = (-11 - 10)/(-3) = (-21)/(-3) = 7`

$\Rightarrow \begin{cases} \dfrac{x-1}{2}=7 \to x - 1 = 14 \to x = 15\\ \dfrac{3-y}{3}=7 \to 3-y = 21 \to y = -18\\\dfrac{5+z}{4}=7 \to 5+z=28 \to z = 23\end{cases}$

Vậy `x=15;y=-18;z=23`

h) Đặt `x/5 = y/7 = k`

`=> x=5k;y=7k`

`x.y = 35 => 5k.7k = 35`

`=> 35k^2= 35`

`=> k^2 = 1`

$\Rightarrow \left[\begin{array}{l} k=1 \to \begin{cases} x=5k=5.1=5\\y=7k=7.1=7\end{cases}\\k=-1 \to \begin{cases} x = 5k=5.(-1)=-5\\y=7k=7.(-1)=-7\end{cases}\end{array}\right.$

Vậy.............

Đáp án:

$\begin{array}{l}
a)\dfrac{x}{y} = \dfrac{{ - 5}}{7}\\
 \Rightarrow \dfrac{x}{{ - 5}} = \dfrac{y}{7} = \dfrac{{x - y}}{{ - 5 - 7}} = \dfrac{{ - 84}}{{ - 12}} = 7\\
 \Rightarrow \left\{ \begin{array}{l}
x = 7.\left( { - 5} \right) =  - 35\\
y = 7.7 = 49
\end{array} \right.\\
c)\dfrac{x}{2} = \dfrac{y}{5} = \dfrac{z}{4}\\
 = \dfrac{{2x}}{4} = \dfrac{{3y}}{{15}} = \dfrac{{2x - 3y + z}}{{4 - 15 + 4}} = \dfrac{{45}}{{ - 7}}\\
 \Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{ - 45}}{7}.2 = \dfrac{{ - 90}}{7}\\
y = \dfrac{{ - 45}}{7}.5 = \dfrac{{ - 225}}{7}\\
z = \dfrac{{ - 45}}{7}.4 = \dfrac{{ - 180}}{7}
\end{array} \right.\\
e)\dfrac{x}{2} = \dfrac{y}{3};\dfrac{y}{5} = \dfrac{z}{4}\\
 \Rightarrow y = \dfrac{{3x}}{2};y = \dfrac{{5z}}{4}\\
 \Rightarrow \dfrac{{3x}}{2} = y = \dfrac{{5z}}{4}\\
 \Rightarrow \dfrac{x}{{10}} = \dfrac{y}{{15}} = \dfrac{z}{{12}}\\
 = \dfrac{{2x}}{{20}} = \dfrac{{3y}}{{45}} = \dfrac{{4z}}{{48}}\\
 = \dfrac{{2x - 3y + 4z}}{{20 - 45 + 48}} = \dfrac{{69}}{{23}} = 3\\
 \Rightarrow \left\{ \begin{array}{l}
x = 30\\
y = 45\\
z = 36
\end{array} \right.\\
g)\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}\\
 \Rightarrow {\left( {\dfrac{x}{2}} \right)^2} = {\left( {\dfrac{y}{3}} \right)^2} = {\left( {\dfrac{z}{4}} \right)^2}\\
 \Rightarrow \dfrac{{{x^2}}}{4} = \dfrac{{{y^2}}}{9} = \dfrac{{{z^2}}}{{16}} = \dfrac{{3{y^2}}}{{27}}\\
 = \dfrac{{{x^2} + 3{y^2} - {z^2}}}{{4 + 27 - 16}} = \dfrac{{60}}{{15}} = 4\\
 \Rightarrow \left\{ \begin{array}{l}
{x^2} = 16\\
{y^2} = 36\\
{z^2} = 64
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x =  \pm 4\\
y =  \pm 6\\
z =  \pm 8
\end{array} \right.\\
b)3x =  - 4y\\
 \Rightarrow \dfrac{x}{{ - 4}} = \dfrac{y}{3} = \dfrac{{2x}}{{ - 8}}\\
 = \dfrac{{2x - y}}{{ - 8 - 3}} = \dfrac{{33}}{{ - 11}} =  - 3\\
 \Rightarrow \left\{ \begin{array}{l}
x = 12\\
y =  - 9
\end{array} \right.\\
d)x:y:z = 3:5:\left( { - 2} \right)\\
 \Rightarrow \dfrac{x}{3} = \dfrac{y}{5} = \dfrac{z}{{ - 2}}\\
 = \dfrac{{5x}}{{15}} = \dfrac{{3z}}{{ - 6}} = \dfrac{{5x - y + 3z}}{{15 - 5 + \left( { - 6} \right)}} = \dfrac{{124}}{4} = 31\\
 \Rightarrow \left\{ \begin{array}{l}
x = 93\\
y = 155\\
z =  - 62
\end{array} \right.\\
f)\dfrac{{x - 1}}{2} = \dfrac{{3 - y}}{3} = \dfrac{{5 + z}}{4}\\
 = \dfrac{{2x - 2}}{4}\\
 = \dfrac{{2x - 2 - \left( {3 - y} \right) - 5 - z}}{{4 - 3 - 4}}\\
 = \dfrac{{2x - 2 - 3 + y - 5 - z}}{{ - 3}}\\
 = \dfrac{{\left( {2x + y - z} \right) - 10}}{{ - 3}}\\
 = \dfrac{{ - 11 - 10}}{{ - 3}}\\
 = \dfrac{{ - 21}}{{ - 3}} = 7
\end{array}$

$\begin{array}{l}
\left\{ \begin{array}{l}
x - 1 = 14\\
3 - y = 21\\
5 + z = 28
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 15\\
y =  - 18\\
z = 23
\end{array} \right.\\
h)\dfrac{x}{5} = \dfrac{y}{7}\\
 \Rightarrow x = \dfrac{{5y}}{7}\\
Thay\,x = \dfrac{{5y}}{7}\,vao:x.y = 35\\
 \Rightarrow \dfrac{{5y}}{7}.y = 35\\
 \Rightarrow \dfrac{5}{7}{y^2} = 35\\
 \Rightarrow {y^2} = 35:\dfrac{5}{7}\\
 \Rightarrow {y^2} = 49\\
 \Rightarrow \left[ \begin{array}{l}
y = 7\\
y =  - 7
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{5}{7}y = 5\\
x = \dfrac{5}{7}y =  - 5
\end{array} \right.
\end{array}$