Question

Đã trả lời bởi chuyên gia

Answer 1

$c, T a c ó : 2 x = 3 y = > \frac{x}{3} = \frac{y}{2} Á p d ụ n g t í n h c h ấ t d ã y t ỉ s ố b ằ n g n h a u t a c ó : \frac{x}{3} = \frac{y}{2} = \frac{y - x}{2 - 3} = \frac{- 5}{- 1} = 5 = > x = 3.5 = 15; y = 2.5 = 10 V ậ y (x; y) = (15; 10)$

$d, T a c ó : x : y : z = 2 : 3 : 5 = > \frac{x}{2} = \frac{y}{3} = \frac{z}{5} Á p d ụ n g t í n h c h ấ t d ã y t ỉ s ố b ằ n g n h a u t a c ó : \frac{x}{2} = \frac{y}{3} = \frac{z}{5} = \frac{- x + y + 2 z}{- 2 + 3 + 5.2} = \frac{22}{11} = 2 = > x = 2.2 = 4; y = 3.2 = 6; z = 5.2 = 10 V ậ y (x; y; z) = (4; 6; 10)$ .

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Answer 2

$c, T a c ó : 2 x = 3 y = > \frac{x}{3} = \frac{y}{2} Á p d ụ n g t í n h c h ấ t d ã y t ỉ s ố b ằ n g n h a u t a c ó : \frac{x}{3} = \frac{y}{2} = \frac{y - x}{2 - 3} = \frac{- 5}{- 1} = 5 = > x = 3.5 = 15; y = 2.5 = 10 V ậ y (x; y) = (15; 10)$

$d, T a c ó : x : y : z = 2 : 3 : 5 = > \frac{x}{2} = \frac{y}{3} = \frac{z}{5} Á p d ụ n g t í n h c h ấ t d ã y t ỉ s ố b ằ n g n h a u t a c ó : \frac{x}{2} = \frac{y}{3} = \frac{z}{5} = \frac{- x + y + 2 z}{- 2 + 3 + 5.2} = \frac{22}{11} = 2 = > x = 2.2 = 4; y = 3.2 = 6; z = 5.2 = 10 V ậ y (x; y; z) = (4; 6; 10)$ .

Answer 3

\[\begin{array}{l}
c){\rm{ }}2x{\rm{ }} = {\rm{ }}3y\\
= > \frac{x}{3} = \frac{y}{2}\\
{\rm{ }}y{\rm{ }}--{\rm{ }}x{\rm{ }} = {\rm{ }} - {\rm{ }}5
\end{array}\]

theo tc dãy tỉ số = nhau:

\[\begin{array}{l}
\frac{x}{3} = \frac{y}{2} = \frac{{y - x}}{{2 - 3}} = \frac{{ - 5}}{{ - 1}} = 5\\
= > \frac{x}{3} = 5 = > x = 15\\
\frac{y}{2} = 5 = > y = 10\\
vay:\left( {x;y} \right) = \left( {15;10} \right)
\end{array}\]

\[\begin{array}{l}
d){\rm{ }}x{\rm{ }}:{\rm{ }}y{\rm{ }}:{\rm{ }}z{\rm{ }} = {\rm{ }}2{\rm{ }}:{\rm{ }}3{\rm{ }}:{\rm{ }}5{\rm{ }}\\
{\rm{ = > }}\frac{x}{2} = \frac{y}{3} = \frac{z}{5}\\
= > \frac{x}{2} = \frac{y}{3} = \frac{{2z}}{{10}}\\
{\rm{ }}2z{\rm{ }} - {\rm{ }}x{\rm{ }} + {\rm{ }}y{\rm{ }} = {\rm{ }}22
\end{array}\]

theo tc dãy tỉ số = nhau:

\[\begin{array}{l}
\frac{x}{2} = \frac{y}{3} = \frac{{2z}}{{10}} = \frac{{ - x + y + 2z}}{{ - 2 + 3 + 10}} = \frac{{22}}{{11}} = 2\\
= > \frac{x}{2} = 2 = > x = 4\\
\frac{y}{3} = 2 = > y = 6\\
\frac{{2z}}{{10}} = 2 = > z = 10\\
vay:\left( {x;y;z} \right) = \left( {4;6;10} \right)
\end{array}\]

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Answer 4

\[\begin{array}{l}
c){\rm{ }}2x{\rm{ }} = {\rm{ }}3y\\
= > \frac{x}{3} = \frac{y}{2}\\
{\rm{ }}y{\rm{ }}--{\rm{ }}x{\rm{ }} = {\rm{ }} - {\rm{ }}5
\end{array}\]

theo tc dãy tỉ số = nhau:

\[\begin{array}{l}
\frac{x}{3} = \frac{y}{2} = \frac{{y - x}}{{2 - 3}} = \frac{{ - 5}}{{ - 1}} = 5\\
= > \frac{x}{3} = 5 = > x = 15\\
\frac{y}{2} = 5 = > y = 10\\
vay:\left( {x;y} \right) = \left( {15;10} \right)
\end{array}\]

\[\begin{array}{l}
d){\rm{ }}x{\rm{ }}:{\rm{ }}y{\rm{ }}:{\rm{ }}z{\rm{ }} = {\rm{ }}2{\rm{ }}:{\rm{ }}3{\rm{ }}:{\rm{ }}5{\rm{ }}\\
{\rm{ = > }}\frac{x}{2} = \frac{y}{3} = \frac{z}{5}\\
= > \frac{x}{2} = \frac{y}{3} = \frac{{2z}}{{10}}\\
{\rm{ }}2z{\rm{ }} - {\rm{ }}x{\rm{ }} + {\rm{ }}y{\rm{ }} = {\rm{ }}22
\end{array}\]

theo tc dãy tỉ số = nhau:

\[\begin{array}{l}
\frac{x}{2} = \frac{y}{3} = \frac{{2z}}{{10}} = \frac{{ - x + y + 2z}}{{ - 2 + 3 + 10}} = \frac{{22}}{{11}} = 2\\
= > \frac{x}{2} = 2 = > x = 4\\
\frac{y}{3} = 2 = > y = 6\\
\frac{{2z}}{{10}} = 2 = > z = 10\\
vay:\left( {x;y;z} \right) = \left( {4;6;10} \right)
\end{array}\]

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