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a) \(x=\dfrac{-2}{7}+\dfrac{9}{7}=1\)
b) \(\dfrac{x}{3}=\dfrac{2}{5}+\dfrac{-4}{3}\)
\(\dfrac{x}{3}=\dfrac{-14}{15}\)
\(\Rightarrow x=\dfrac{3.-14}{15}=\dfrac{-14}{5}\)
\(x=\dfrac{-2}{7}+\dfrac{9}{7}\)
\(x=1\)
\(\dfrac{-5}{9}\)=\(\dfrac{-45}{b}\)
⇒ b= [9. (-45)] : -5
⇒ b= -405 : -5
⇒ b= 81
⇒ \(\dfrac{-45}{81}\)
\(\dfrac{a}{27}\)= \(\dfrac{-5}{9}\)
⇒ a= [ 27 .(-5) ] : 9
⇒ a= -135 : 9
⇒ a= -15
⇒ \(\dfrac{-15}{27}\)
⇒ \(\dfrac{-15}{27}\)=\(\dfrac{-5}{9}\)=\(\dfrac{-45}{81}\)
\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
⇒\(a=\dfrac{-5.27}{9}=-15\)
⇒\(b=\dfrac{-45.9}{5}=-81\)
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
\(\dfrac{x}{9}-\dfrac{3}{y}=\dfrac{1}{18}\left(ĐKXĐ:y\ne0\right)\)
\(\Rightarrow\dfrac{xy-27}{9y}=\dfrac{1}{18}\)
\(\Rightarrow18\left(xy-27\right)=9y\)
\(\Rightarrow2\left(xy-27\right)=y\)
\(\Rightarrow2xy-54=y\)
\(\Rightarrow2xy-y=54\Rightarrow y\left(2x-1\right)=54\)
\(\Rightarrow y=\dfrac{54}{2x-1}\)
- 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 : \(\left(x;y\right)\in\left\{\left(1;54\right),\left(0;-54\right),\left(2;18\right),\left(-1;-18\right),\left(5;6\right),\left(-4;-6\right),\left(14;2\right),\left(-13;-2\right)\right\}\)
\(\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{3}{9}\)
\(\Leftrightarrow\dfrac{x}{27}=-\dfrac{1}{9}\)
\(\Leftrightarrow9x=-27\)
\(\Leftrightarrow x=-\dfrac{27}{9}\)
\(\Leftrightarrow x=-3\)
=>x/27=1/9+3/27=1/9+1/9=2/9
=>x=6
`x/27+(-3/27)=1/9`
`=>x/27+(-1/9)=1/9`
`=>x/27=1/9-(-1/9)`
`=>x/27=1/9+1/9`
`=>x/27=2/9`
`=>x/27=6/27`
`=>x=6`
Vậy `x=6`