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\(\dfrac{x}{9}=\dfrac{3}{y}+\dfrac{1}{18}\left(y\ne0\right)\)
\(\Rightarrow\dfrac{2xy}{18y}=\dfrac{54}{18y}+\dfrac{y}{18y}\)
\(\Rightarrow2xy=54+y\)
\(\Rightarrow2xy-y=54\)
\(\Rightarrow xy-\dfrac{y}{2}=27\)
\(\Rightarrow y\left(x-\dfrac{1}{2}\right)=27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right);y\in\left\{1;3;9;27\right\}\)
\(\Rightarrow\left(x;\right)y\in\left\{\left(\dfrac{1}{2};27\right);\left(\dfrac{5}{2};9\right);\left(\dfrac{17}{2};3\right);\left(\dfrac{53}{2};1\right)\right\}\)
\(\Rightarrow\left(x;y\right)\in\varnothing\left(x;y\inℕ\right)\)
\(\dfrac{x}{9}-\dfrac{3}{y}=\dfrac{1}{18}\left(ĐKXĐ:y\ne0\right)\)
\(\Rightarrow\dfrac{xy-27}{9y}=\dfrac{1}{18}\)
\(\Rightarrow18\left(xy-27\right)=9y\)
\(\Rightarrow2\left(xy-27\right)=y\)
\(\Rightarrow2xy-54=y\)
\(\Rightarrow2xy-y=54\Rightarrow y\left(2x-1\right)=54\)
\(\Rightarrow y=\dfrac{54}{2x-1}\)
- Suy ra 54 chia hết cho 2x - 1
\(\Rightarrow2x-1\inƯ\left(54\right)\)
\(\Rightarrow2x-1\in\left\{1;-1;2;-2;3;-3;9;-9;27;-27\right\}\)
Cho 2x - 1 bằng từng giá trị ở trên, ta tìm được :
\(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2};2;-1;5;-4;14;-13\right\}\). Mà x không có giá trị ngoài tập số nguyên.
\(\Rightarrow x\in\left\{-13;-4;-1;0;1;2;5;14\right\}\)
Thay các giá trị x trên vừa tìm được vào y :
\(\Rightarrow y\in\left\{54;-54;18;-18;6;-6;2;-2\right\}\)
Vậy : Các số x và y thỏa mãn đề bài là : \(\left(x;y\right)\in\left\{\left(1;54\right),\left(0;-54\right),\left(2;18\right),\left(-1;-18\right),\left(5;6\right),\left(-4;-6\right),\left(14;2\right),\left(-13;-2\right)\right\}\)
b)3x+1/18+2y/12=2/9 và x-y=1
2(3x+1)/18x2+2y x 3/12x3=2x4/9x4
6x+2+6y=8
6x+6y=8-2=6
6(x+y)=6
x+y=6:6=1(1)
theo đề bài ta có:x-y=1 suy ra x=y+1
thay x=y+1 vào (1)
y+1+y=1
2y=1-1=0
y=0:2=0
x=0+1=1
xong rồi câu a) ko biết làm
a) <=> \(\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y+2}\Leftrightarrow x-1+2=\dfrac{9}{y+2}\)
\(\Leftrightarrow x=\dfrac{9}{y+2}-1\) với mỗi giá trị của y khác -2 luôn tìm được x
từ và x-y =1 áp cho cả câu (a) thì
\(x-y=1=>x+1=y+2\)
\(y+2=\dfrac{9}{y+2}\Leftrightarrow\left\{{}\begin{matrix}y\ne-2\\\left(y+2\right)^2=9\end{matrix}\right.\)
y+2 = 3 => y = 1 =>x=2
y+2 =-3 => y =-5=> x=-4
a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)
\(\Rightarrow-4< x< \dfrac{-3}{10}\)
\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)
\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)
b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)
\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)
\(\Rightarrow x=\varnothing\)
c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)
\(\Rightarrow x\in\left\{1;2\right\}\)
+) Với \(x=1\)
\(\Rightarrow y\in\left\{1;2\right\}\)
+) Với \(x=2\)
\(\Rightarrow y=2\)
Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).
\(\Leftrightarrow\left(11+\dfrac{7}{18}-9-\dfrac{13}{18}\right):x=\dfrac{5}{3}+\dfrac{35}{33}\cdot\dfrac{11}{7}=\dfrac{10}{3}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{10}{3}=\dfrac{1}{2}\)
\(\dfrac{x}{9}-\dfrac{3}{y}=\dfrac{1}{18}\)
\(\dfrac{x}{9}-\dfrac{1}{18}=\dfrac{3}{y}\)
\(\dfrac{2x}{18}-\dfrac{1}{18}=\dfrac{3}{y}\)
\(\dfrac{2x-1}{18}=\dfrac{3}{y}\)
\(\Rightarrow\)(2x-1).y=18.3=54
54 có các ước là: \(\pm1;\pm2;\pm3;\pm6;\pm9;\pm18;\pm27;\pm54\)
*\(\left\{{}\begin{matrix}2x-1=1\\y=54\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=2\\y=54\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=54\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}2x-1=54\\y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=55\\y=1\end{matrix}\right.\)\(\notin\) N ( Loại)
*\(\left\{{}\begin{matrix}2x-1=2\\y=27\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=3\\y=27\end{matrix}\right.\) \(\notin\) N ( Loại )
*\(\left\{{}\begin{matrix}2x-1=27\\y=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=28\\y=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=14\\y=2\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}2x-1=3\\y=18\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=4\\y=18\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=18\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}2x-1=18\\y=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=19\\y=3\end{matrix}\right.\)\(\notin\) N ( Loại )
*\(\left\{{}\begin{matrix}2x-1=6\\y=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=7\\y=9\end{matrix}\right.\)\(\notin\) N ( Loại )
*\(\left\{{}\begin{matrix}2x-1=9\\y=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=10\\y=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\y=6\end{matrix}\right.\)
Vậy có các cặp (x,y) t/m đề bài là : (1,54) ; (14,2) ; (2,18) ; (5,6)
\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
