Giải thích các bước giải:

a) $2x=3y \Rightarrow \dfrac{x}{3}=\dfrac{y}{2} \Rightarrow \dfrac{x}{21}=\dfrac{y}{14} (1)\\5y = 7z \Rightarrow \dfrac{y}{7}=\dfrac{z}{5} \Rightarrow \dfrac{y}{14}=\dfrac{z}{10} (2)$

Từ (1) và (2) `=> x/21 = y/14 = z/10`

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

$\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x-7y+5z}{3.21-7.14+5.10}=\dfrac{-16}{15}\\ \Rightarrow \begin{cases} \dfrac{x}{21}=\dfrac{-16}{15} \rightarrow x = \dfrac{-112}{5}\\ \dfrac{y}{14}=\dfrac{-16}{15} \rightarrow y=\dfrac{-224}{15}\\ \dfrac{z}{10}=\dfrac{-16}{15} \rightarrow z = \dfrac{-32}{3}\\\end{cases}$

Vậy `x = (-112)/5; y = (-224)/15; z = (-32)/3`

b) `4x = 7y => x/7 = y/4 => x^2/49 = y^2/16`

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

$\dfrac{x^2}{49}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{49+16}=\dfrac{260}{65}=4\\ \Rightarrow \begin{cases} \dfrac{x^2}{49}=4 \rightarrow x^2 = 196 \rightarrow x = \pm 14\\ \dfrac{y^2}{16}=4 \rightarrow y^2=64 \rightarrow y = \pm 8\\\end{cases}$

Vậy `(x;y) = (14;8);(-14;-8)`

c) Đặt `x/2 = y/4 = k`

$\Rightarrow \begin{cases} x = 2k\\y=4k\\\end{cases}\\x^2.y^2=4 \Rightarrow (2k)^2.(4k)^2=4\\ \Rightarrow 4k^2.16k^2=4\\ \Rightarrow 64k^4 = 4 \Rightarrow k^4= \dfrac{1}{16}\\ \Rightarrow k^4 = \bigg(\pm \dfrac{1}{2}\bigg)^4 \\\Rightarrow \left[\begin{array}{l} k = \dfrac{1}{2} \rightarrow \begin{cases} x=2k =2 . \dfrac{1}{2}=1\\ y=4k =4 . \dfrac{1}{2}=2\\\end{cases}\\ k = \dfrac{-1}{2} \rightarrow \begin{cases} x = 2k = 2 . \dfrac{-1}{2}=-1\\y = 4k = 4 . \dfrac{-1}{2}=-2\\\end{cases}\end{array}\right.$

Vậy `(x;y) = (1;2);(-1;-2)`

d) `x:y:z = 4 : 5:6 => x/4=y/5=z/6`

`⇒ x^2/16 = y^2/25 = z^2/36`

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

$\dfrac{x^2}{16}=\dfrac{y^2}{25}=\dfrac{z^2}{36}=\dfrac{x^2-2y^2+z^2}{16-2.25+36}=\dfrac{18}{2}=9\\ \Rightarrow \begin{cases} \dfrac{x^2}{16}=9 \rightarrow x^2=144 \rightarrow x = \pm 12\\ \dfrac{y^2}{25}=9 \rightarrow y^2=225 \rightarrow y = \pm 15\\ \dfrac{z^2}{36} = 9 \rightarrow z = 324 \rightarrow z = \pm 18\\\end{cases}$

Vậy `(x;y;z) = (12;15;18);(-12;-15;-18)`

e) $0,25x:3=\dfrac{5}{6}:0,125\\ ⇒0,25x :3= \dfrac{5}{6}:\dfrac{1}{8}\\⇒ 0,25x=\dfrac{5}{6}.8.3\\⇒ 0,25x=20\\⇒ x = 20:0,25\\⇒x=80 $

f) $3,8:2x=\dfrac{1}{4}:\dfrac{8}{3}\\⇒ \dfrac{19}{5}:2x=\dfrac{1}{4}.\dfrac{3}{8}\\⇒ \dfrac{19}{5}:2x=\dfrac{3}{32}\\⇒2x = \dfrac{19}{5}:\dfrac{3}{32}\\⇒2x=\dfrac{19}{5}.\dfrac{32}{3}\\⇒2x=\dfrac{608}{15}\\⇒x=\dfrac{304}{15}$

Đáp án:

f. \(x = \dfrac{{76}}{5}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\left\{ \begin{array}{l}
x = \dfrac{3}{2}y\\
z = \dfrac{5}{7}y\\
3.\dfrac{3}{2}y - 7y + 5.\dfrac{5}{7}y =  - 16\left( * \right)
\end{array} \right.\\
\left( * \right) \to \left( {\dfrac{9}{2} - 7 + \dfrac{{25}}{7}} \right)y =  - 16\\
 \to y =  - \dfrac{{224}}{{15}}\\
 \to x =  - \dfrac{{112}}{5};z =  - \dfrac{{32}}{3}\\
b.\left\{ \begin{array}{l}
x = \dfrac{7}{4}y\\
{\left( {\dfrac{7}{4}y} \right)^2} + {y^2} = 260
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{7}{4}y\\
\dfrac{{65}}{{16}}{y^2} = 260
\end{array} \right.\\
 \to \left[ \begin{array}{l}
y = 64\\
y =  - 64
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 112\\
x =  - 112
\end{array} \right.\\
c.\left\{ \begin{array}{l}
x = \dfrac{1}{2}y\\
\dfrac{1}{4}{y^2}.{y^2} = 4
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
{y^4} = 16\\
x = \dfrac{1}{2}y
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
{y^2} = 4\\
x = \dfrac{1}{2}y
\end{array} \right.\\
 \to \left[ \begin{array}{l}
y = 2\\
y =  - 2
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\\
x =  - 1
\end{array} \right.\\
e.\dfrac{1}{4}.x:3 = \dfrac{5}{6}:\dfrac{1}{8}\\
 \to \dfrac{x}{{12}} = \dfrac{{40}}{6}\\
 \to x = 80\\
f.\dfrac{{19}}{5}:2x = \dfrac{1}{4}:\dfrac{8}{3}\\
 \to \dfrac{{19}}{{10x}} = \dfrac{3}{{24}}\\
 \to 30x = 19.24\\
 \to x = \dfrac{{76}}{5}
\end{array}\)