Toán Lớp 7: Tìm x,y biết:
a) 2x=5y và 3x+y=1
b) x/2 = y/3 và x ² +2xy =16
c) x/2 = -3/5 · y và x ³ + y ³ = 91
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Toán Lớp 7: Tìm x,y biết:
a) 2x=5y và 3x+y=1
b) x/2 = y/3 và x ² +2xy =16
c) x/2 = -3/5 · y và x ³ + y ³ = 91
Comments ( 1 )
a)2x = 5y\\
\Leftrightarrow \dfrac{x}{5} = \dfrac{y}{2} = \dfrac{{3x}}{{15}} = \dfrac{{3x + y}}{{15 + 2}} = \dfrac{1}{{17}}\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{{17}}.5 = \dfrac{5}{{17}}\\
y = \dfrac{1}{{17}}.2 = \dfrac{2}{{17}}
\end{array} \right.\\
Vậy\,x = \dfrac{5}{{17}};y = \dfrac{2}{{17}}\\
b)\dfrac{x}{2} = \dfrac{y}{3} = k \Leftrightarrow \left\{ \begin{array}{l}
x = 2k\\
y = 3k
\end{array} \right.\left( {k \ne 0} \right)\\
{x^2} + 2xy = 16\\
\Leftrightarrow {\left( {2k} \right)^2} + 2.2k.3k = 16\\
\Leftrightarrow 4{k^2} + 12{k^2} = 16\\
\Leftrightarrow 16{k^2} = 16\\
\Leftrightarrow {k^2} = 1\\
\Leftrightarrow \left[ \begin{array}{l}
k = 1 \Leftrightarrow x = 2;y = 3\\
k = – 1 \Leftrightarrow x = – 2;y = – 3
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {2;3} \right);\left( { – 2; – 3} \right)} \right\}\\
c)\dfrac{x}{2} = \dfrac{{ – 3}}{5}\\
\Leftrightarrow x = – \dfrac{6}{5}\\
{x^3} + {y^3} = 91\\
\Leftrightarrow {\left( { – \dfrac{6}{5}} \right)^3} + {y^3} = 91\\
\Leftrightarrow {y^3} = \dfrac{{11591}}{{125}}\\
\Leftrightarrow y = \dfrac{{\sqrt[3]{{11591}}}}{5}\\
Vậy\,x = \dfrac{{ – 6}}{5};y = \dfrac{{\sqrt[3]{{11591}}}}{5}
\end{array}$