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b) Ta có: \(x\in N\)
Do đó, để \(\left(3x-2\right)\left(2y-3\right)=1\)
\(\Rightarrow\orbr{\begin{cases}3x-2=1\Rightarrow3x=3\Rightarrow x=1\\2y-3=1\Rightarrow2y=4\Rightarrow y=2\end{cases}}\)
b) ( 3x - 2 ) ( 2y - 3 ) = 1
=>\(Vì1.1=1=>\orbr{\begin{cases}3x-2=1\\2y-3=1\end{cases}}\)
\(=>\orbr{\begin{cases}x=1\\y=2\end{cases}}\)
Vậy (x,y)=(1;2)
Bài b làm tương tự nhé chỉ có xét nhiều hơn thôi
khó vãi
a, Giải:
Ta có: \(\dfrac{x}{5}=\dfrac{y}{7}\Rightarrow\dfrac{7}{5}x=y\)
\(x-y=-10\)
\(\Rightarrow x-\dfrac{7}{5}x=-10\)
\(\Rightarrow\dfrac{-2}{5}x=-10\)
\(\Rightarrow x=25\Rightarrow y=35\)
Vậy x = 25, y = 35
b, Giải:
Ta có: \(3x=4y\Rightarrow\dfrac{3}{4}x=y\)
\(y+x=14\)
\(\Rightarrow\dfrac{3}{4}x+x=14\)
\(\Rightarrow\dfrac{7}{4}x=14\)
\(\Rightarrow x=8\Rightarrow y=6\)
Vậy x = 8, y = 6
c, Ta có: \(\dfrac{4}{x}=\dfrac{2}{y}\Rightarrow\dfrac{x}{4}=\dfrac{y}{2}\Rightarrow x=2y\)
\(2x-y=12\)
\(\Rightarrow4y-y=12\)
\(\Rightarrow3y=12\)
\(\Rightarrow y=4\)
\(\Rightarrow x=8\)
Vậy x = 8, y = 4
a) Ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=7k\end{matrix}\right.\)
Do \(x-y=-10\Leftrightarrow5k-7k=-10\)
\(\Leftrightarrow\left(5-7\right)k=-10\)
\(\Leftrightarrow\left(-2\right)k=-10\)
\(\Leftrightarrow k=\left(-10\right):\left(-2\right)=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=5.5=25\\y=7k=7.5=35\end{matrix}\right.\)
b) Xét \(3x=4y\Leftrightarrow\dfrac{x}{4}=\dfrac{y}{3}=m\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4m\\y=3m\end{matrix}\right.\)
Do \(y+x=14\Leftrightarrow3m+4m=14\)
\(\Leftrightarrow\left(3+4\right)m=14\)
\(\Leftrightarrow7m=14\)
\(\Leftrightarrow m=14:7=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=4m=4.2=8\\y=3m=3.2=6\end{matrix}\right.\)
c) Ta có \(\dfrac{4}{x}=\dfrac{2}{y}=n\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{n}\\y=\dfrac{2}{n}\end{matrix}\right.\)
Do \(2x-y=12\Leftrightarrow2.\dfrac{4}{n}-\dfrac{2}{n}=12\)
\(\Leftrightarrow\dfrac{8}{n}-\dfrac{2}{n}=12\)
\(\Leftrightarrow\dfrac{6}{n}=12\)
\(\Leftrightarrow n=\dfrac{6}{12}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{n}=\dfrac{4}{\dfrac{1}{2}}=8\\y=\dfrac{2}{n}=\dfrac{2}{\dfrac{1}{2}}=4\end{matrix}\right.\)
Câu 3:(x+1).(y-7)=13
Th1:x+1=13
x=13-1
x=12
Th2:y-7=13
y =13+7
y=20
Vậy x=12 hoặc y=20
\(A=\left|x+19\right|+\left|y-5\right|+2020\)
Ta có : \(\left|x+19\right|\ge0\forall x;\left|y-5\right|\ge0\forall y;2020>0\)
Suy ra : \(\left|x+19\right|+\left|y-5\right|+2020\ge2020\)
Dấu ''='' xảy ra : \(\hept{\begin{cases}x+19=0\\y-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-19\\y=5\end{cases}}}\)
Vậy GTNN A = 2020 <=> x = -19 ; y = 5
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 \(y=8\).
a) Ta có :
\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{4}{x}\)
\(\Leftrightarrow\dfrac{5}{6}-\dfrac{2y}{6}=\dfrac{4}{x}\)
\(\Leftrightarrow\dfrac{5-2y}{6}=\dfrac{4}{x}\)
\(\Leftrightarrow\left(5-2y\right)x=6.4=24\)
Vì \(x,y\in N\Leftrightarrow5-2y\in N;5-2y;x\inƯ\left(24\right)\)
Ta có bảng :
| \(x\) | \(y\) | \(5-2y\) | \(Đk\) \(x,y\in N\) |
| \(1\) | \(\dfrac{-19}{2}\) | \(24\) | loại |
| \(2\) | \(\dfrac{-7}{2}\) | \(12\) | loại |
| \(3\) | \(\dfrac{-3}{2}\) | 2\(8\) | loại |
| \(4\) | \(\dfrac{1}{2}\) | \(6\) | loại |
| \(8\) | \(1\) | \(3\) | thỏa mãn |
| \(12\) | \(\dfrac{3}{2}\) | \(2\) | loại |
| \(24\) | \(2\) | \(1\) | thỏa mãn |
Vậy ...
\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{y}{3}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{2y}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5-2y}{6}\)
\(\Rightarrow x\left(5-2y\right)=24\)
\(\Rightarrow x;5-2y\inƯ\left(24\right)\)
Xét ước là xong
\(3x-xy-4y+12=17\)
\(\Rightarrow x\left(3-y\right)+4\left(3-y\right)=17\)
\(\Rightarrow\left(x+4\right)\left(3-y\right)=17\)
\(\Rightarrow x+4;3-y\inƯ\left(17\right)\)
\(Ư\left(17\right)=\left\{\pm1;\pm17\right\}\)
Xét ước
Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
a)\(3x=4y\Rightarrow x=\dfrac{4}{3}y\)
\(x+y=-14\)
\(\Rightarrow x=-14:\left(4+3\right).4=-8\)
\(y=-14--8=-6\)
b)\(\dfrac{4}{x}=\dfrac{2}{y}\Rightarrow4y=2x\)
\(2x-y=12\Rightarrow2x=y+12\)
\(y+12=4y\)
\(12=3y\)
\(y=4\)
\(x=4.4:2=8\)
bạn còn có cách làm khác ko