TK(/hoi-dap/detail/40878446416.html) 

Có a,b = b,a x  3 + 1,3

=> ab = ba x 3 + 1,3

=> 10a + b = 30b + 3a + 13

=> 7a - 29b = 13

=> 7a = 29b + 13

Mà \(b>0=>29b+13>13\)

=> 7a > 13 

=> a \(\ge2\)

TH1: a = 2 => b = \(\dfrac{1}{29}\left(l\right)\)

TH2: a = 3 => b = \(\dfrac{8}{29}\left(l\right)\)

TH3: a = 4 => b = \(\dfrac{15}{29}\left(l\right)\)

TH4: a = 5 => \(b=\dfrac{22}{29}\) (l)

TH5: a = 6 => \(b=1\) (thỏa mãn)

TH6: a = 7 => \(b=\dfrac{36}{29}\left(l\right)\)

TH7: a = 8 => b = \(\dfrac{43}{29}\left(l\right)\)

TH8: a = 9 => \(b=\dfrac{50}{29}\left(l\right)\)

Vậy a = 6, b = 1

ab = 61

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a)\(\left[{}\begin{matrix}\dfrac{5}{2}-x=\dfrac{1}{3}\\\dfrac{5}{2}-x=-\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=\dfrac{17}{6}\end{matrix}\right.\)

b) 8/6-x-1/5=0

9/6-x=1/5

x=13/10

a = 1,2

b = 1,8

a) \(\left|2,5-x\right|=1,3\)

\(\Leftrightarrow\left[{}\begin{matrix}2,5-x=1,3\\2,5-x=-1,3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)

b) \(1,6-\left|x-0,2\right|=0\)

\(\Leftrightarrow\left|x-0,2\right|=1,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)

a.25 - y² = 8(x - 2009)

 25 - y² = 8x - 16072

 - 8x = -16072 - 25 + y²

 - 8x = -16097 + y²

 x = /8 - 1/8y²

-Vậy x = /8 - 1/8y²

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