Câu 1 Tìm x,y ∈ z biết
a) $\frac{x}{3}$ = $\frac{-2}{y}$ x>0>y
b) $\frac{5}{y}$ và $\frac{x}{-2}$ y<0
Câu 1 Tìm x,y ∈ z biết
a) $\frac{x}{3}$ = $\frac{-2}{y}$ x>0>y
b) $\frac{5}{y}$ và $\frac{x}{-2}$ y<0
Đáp án:
$\begin{array}{l}
1)a)\dfrac{x}{3} = \dfrac{{ – 2}}{y}\\
\Rightarrow x.y = 3.\left( { – 2} \right) = – 6 = \left( { – 1} \right).6\\
x > 0 > y\\
\Rightarrow \left[ \begin{array}{l}
x = 3;y = – 2\\
x = 2;y = – 3\\
x = 1;y = – 6\\
x = 6;y = – 1
\end{array} \right.\\
b)\dfrac{5}{y} = \dfrac{x}{{ – 2}}\\
\Rightarrow x.y = 5.\left( { – 2} \right) = 10 = 1.\left( { – 10} \right)\\
y < 0 < x\\
\Rightarrow \left[ \begin{array}{l}
x = 5;y = – 2\\
x = 2;y = – 5\\
x = 1;y = – 10\\
x = 10;y = – 1
\end{array} \right.\\
c)\dfrac{{x – 3}}{{y – 2}} = \dfrac{3}{2}\\
\Rightarrow 3.\left( {y – 2} \right) = 2.\left( {x – 3} \right)\\
\Rightarrow 3.y – 3.2 = 2.x – 2.3\\
\Rightarrow 3.y – 6 = 2.x – 6\\
\Rightarrow 3.y = 2.x\\
\Rightarrow y = \dfrac{2}{3}.x\\
Do:x – y = 4\\
\Rightarrow x – \dfrac{2}{3}.x = 4\\
\Rightarrow \dfrac{1}{3}.x = 4 \Rightarrow x = 4.3 = 12\\
\Rightarrow y = \dfrac{2}{3}.x = \dfrac{2}{3}.12 = 8\\
Vay\,x = 12;y = 8\\
C2)a)16 = {2^4}\\
72 = {2^3}{.3^2}\\
\Rightarrow MSC = {2^4}{.3^2} = 144\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{ – 5}}{{16}} = \dfrac{{ – 5.9}}{{144}} = \dfrac{{ – 45}}{{144}}\\
\dfrac{{33}}{{ – 72}} = \dfrac{{ – 33.2}}{{72.2}} = \dfrac{{ – 66}}{{144}}
\end{array} \right.\\
b)56 = {2^3}.7;63 = {3^2}.7;144 = {2^4}{.3^2}\\
\Rightarrow MSC = {2^4}{.3^2}.7 = 1008\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{ – 1}}{{56}} = \dfrac{{ – {{2.3}^2}}}{{1008}} = \dfrac{{ – 18}}{{1008}}\\
\dfrac{{ – 11}}{{ – 63}} = \dfrac{{11.16}}{{1008}} = \dfrac{{176}}{{1008}}\\
\dfrac{{26}}{{144}} = \dfrac{{26.7}}{{1008}} = \dfrac{{182}}{{1008}}
\end{array} \right.\\
c)24 = {2^3}.3;15 = 3.5;40 = {2^3}.5\\
\Rightarrow MSC = {2^3}.3.5 = 120\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{ – 7}}{{24}} = \dfrac{{ – 7.5}}{{120}} = \dfrac{{ – 35}}{{120}}\\
\dfrac{{14}}{{15}} = \dfrac{{14.8}}{{120}} = \dfrac{{112}}{{120}}\\
\dfrac{{ – 3}}{{40}} = \dfrac{{ – 3.3}}{{120}} = \dfrac{{ – 9}}{{120}}
\end{array} \right.
\end{array}$