Toán Lớp 7: Tìm x, y, z biết:
$\frac{y+z+1}{x}$ = $\frac{x+z+2}{y}$ = $\frac{x+y-3}{z}$ = $\frac{1}{x+y+z}$
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Toán Lớp 7: Tìm x, y, z biết:
$\frac{y+z+1}{x}$ = $\frac{x+z+2}{y}$ = $\frac{x+y-3}{z}$ = $\frac{1}{x+y+z}$
Comments ( 1 )
\dfrac{{y + z + 1}}{x} = \dfrac{{x + z + 2}}{y} = \dfrac{{x + y – 3}}{z} = \dfrac{1}{{x + y + z}}\\
= \dfrac{{y + z + 1 + x + z + 2 + x + y – 3}}{{x + y + z}}\\
= \dfrac{{2.\left( {x + y + z} \right)}}{{x + y + z}} = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
y + z + 1 = 2x\\
x + z + 2 = 2y\\
x + y – 3 = 2z\\
x + y + z = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y + z = 2x – 1\\
x + z = 2y – 2\\
x + y = 2z + 3\\
x + y + z = \dfrac{1}{2}
\end{array} \right.\\
+ y + z = 2x – 1\\
\Leftrightarrow x + 2x – 1 = \dfrac{1}{2}\\
\Leftrightarrow 3x = \dfrac{3}{2}\\
\Leftrightarrow x = \dfrac{1}{2}\\
+ x + z = 2y – 2\\
\Leftrightarrow y + 2y – 2 = \dfrac{1}{2}\\
\Leftrightarrow 3y = \dfrac{5}{2}\\
\Leftrightarrow y = \dfrac{5}{6}\\
Do:x + y + z = \dfrac{1}{2}\\
\Leftrightarrow z = \dfrac{1}{2} – x – y = \dfrac{1}{2} – \dfrac{1}{2} – \dfrac{5}{6} = – \dfrac{5}{6}\\
Vậy\,x = \dfrac{1}{2};y = \dfrac{5}{6};z = – \dfrac{5}{6}
\end{array}$