Question

Tìm x , biết :

( \(\dfrac{2}{3}\) + x ) . ( \(\dfrac{3}{5}\) - 2x ) = 10

Giúp mình nha !

Answer 1

\(\left(\dfrac{2}{3}+x\right)\left(\dfrac{3}{5}-2x\right)=10\)

\(\Leftrightarrow\dfrac{2}{3}+x;\dfrac{3}{5}-2x\inƯ\left(10\right)\)

\(Ư\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

\(\Leftrightarrow\dfrac{2}{3}+x=1\Rightarrow x=\dfrac{1}{3}\)

\(\dfrac{3}{5}-2x=10\Rightarrow2x=\dfrac{-47}{5}\Rightarrow x=\dfrac{-47}{10}\)

\(\Leftrightarrow\dfrac{2}{3}+x=-1\Rightarrow x=\dfrac{-5}{3}\)

\(\dfrac{3}{5}-2x=-10\Rightarrow2x=\dfrac{53}{5}\Rightarrow x=\dfrac{53}{10}\)

\(\Leftrightarrow\dfrac{2}{3}+x=2\Rightarrow x=\dfrac{4}{3}\)

\(\dfrac{3}{5}-2x=5\Rightarrow2x=\dfrac{-22}{5}\Rightarrow x=\dfrac{-11}{5}\)

\(\Leftrightarrow\dfrac{2}{3}+x=-2\Rightarrow x=\dfrac{-8}{3}\)

\(\dfrac{3}{5}-2x=-5\Rightarrow2x=\dfrac{28}{5}\Rightarrow x=\dfrac{14}{5}\)

\(\Leftrightarrow\dfrac{2}{3}+x=5\Rightarrow x=\dfrac{13}{3}\)

\(\dfrac{3}{5}-2x=2\Rightarrow2x=\dfrac{-7}{5}\Rightarrow x=\dfrac{-7}{10}\)

\(\Leftrightarrow\dfrac{2}{3}+x=-5\Rightarrow x=\dfrac{-17}{3}\)

\(\dfrac{3}{5}-2x=-2\Rightarrow2x=\dfrac{13}{5}\Rightarrow x=\dfrac{13}{10}\)

\(\Leftrightarrow\dfrac{2}{3}+x=10\Rightarrow x=\dfrac{28}{3}\)

\(\dfrac{3}{5}-2x=1\Rightarrow2x=\dfrac{-2}{5}\Rightarrow x=\dfrac{-1}{5}\)

\(\Leftrightarrow\dfrac{2}{3}+x=-10\Rightarrow x=\dfrac{-32}{3}\)

\(\dfrac{3}{5}-2x=-1\Rightarrow2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\)

Answer 2

Đề có vấn đề, đáng lẽ phải tìm x, y chứ...

\(\left(\dfrac{2}{3}+x\right)\cdot\left(\dfrac{3}{5}-2y\right)=10\)

Ta lập bảng sau:

\(\dfrac{2}{3}+x\)	1	2	5	10	-1	-2	-5	-10
\(x\)	\(\dfrac{1}{3}\)	\(\dfrac{4}{3}\)	\(\dfrac{13}{3}\)	\(\dfrac{28}{3}\)	\(\dfrac{-5}{3}\)	\(\dfrac{-8}{3}\)	\(\dfrac{-17}{3}\)	\(\dfrac{-32}{3}\)
\(\dfrac{3}{5}-2x\)	10	5	2	1	-10	-5	-2	-1
\(y\)	\(\dfrac{-47}{10}\)	\(\dfrac{-11}{5}\)	\(\dfrac{-7}{10}\)	\(\dfrac{-1}{5}\)	\(\dfrac{53}{10}\)	\(\dfrac{14}{5}\)	\(\dfrac{13}{10}\)	\(\dfrac{4}{5}\)

Vậy các cặp (x;y) là: \(\left(\dfrac{1}{3};\dfrac{-47}{10}\right);\left(\dfrac{4}{3};\dfrac{-11}{5}\right);\left(\dfrac{13}{3};\dfrac{-7}{10}\right);\left(\dfrac{28}{3};\dfrac{-1}{5}\right);\left(\dfrac{-5}{3};\dfrac{53}{10}\right);\left(\dfrac{-8}{3};\dfrac{14}{5}\right);\left(\dfrac{-17}{3};\dfrac{13}{10}\right);\left(\dfrac{-32}{3};\dfrac{4}{5}\right).\)