\(\dfrac{x}{9}=\dfrac{3}{y}+\dfrac{1}{18}\left(y\ne0\right)\)

\(\Rightarrow\dfrac{2xy}{18y}=\dfrac{54}{18y}+\dfrac{y}{18y}\)

\(\Rightarrow2xy=54+y\)

\(\Rightarrow2xy-y=54\)

\(\Rightarrow xy-\dfrac{y}{2}=27\)

\(\Rightarrow y\left(x-\dfrac{1}{2}\right)=27\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right);y\in\left\{1;3;9;27\right\}\)

\(\Rightarrow\left(x;\right)y\in\left\{\left(\dfrac{1}{2};27\right);\left(\dfrac{5}{2};9\right);\left(\dfrac{17}{2};3\right);\left(\dfrac{53}{2};1\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\varnothing\left(x;y\inℕ\right)\)

M_min=-1 khi y=3 và x=+-3

Làm như nào vậy. bạn giải rõ ràng ra đi

Ta có: \(\left(x^2-9\right)^2\ge0\forall x\)

\(\left|y-3\right|\ge0\forall y\)

 \(\Rightarrow\left(x^2-9\right)^2+ \left|y-3\right|\ge0\forall x,y\)

 \(\Rightarrow\left(x^2-9\right)^2+\left|y-3\right|-1\ge-1\forall x,y\)

\(\Rightarrow M\ge-1\forall x,y\)

Dấu \("="\) xảy ra khi \(\left(x^2-9\right)^2=0;\left|y-3\right|=0\)

+) \(\left(x^2-9\right)^2=0\Rightarrow x^2-9=0\)

\(\Rightarrow x=+-3\)

+) \(\left|y-3\right|=0\Rightarrow y-3=0\Rightarrow y=3\)

Vậy \(Min_M=-1\) khi \(x=+-3;y=3.\)

\(\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y+2}\)

\(\dfrac{x-1}{9}+\dfrac{3}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x-1+3}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x-\left(1-3\right)}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x-\left(-2\right)}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x+2}{9}=\dfrac{1}{y+2}\)

\(\left(x+2\right)\left(y+2\right)=9\)

=> (X+2) ; (y+2) ϵ Ư(9)

TH1: x+2 = 1 => x = -1

y+2=9 => y = 7

TH2: x+2 = 9 => x = 7

=> y +2 = 1 => y =-1

TH3:x+2 = -9 => x = -11

y+2 = -1 => y=-3

TH4: x+2 = -1 => x =-3

y+2 = -9 => x=-11

TH5: x+2 = -3 => x =-5

y+2 = -3 => y=-5

TH6: x+2 =3 =>  x = 1

y+2=3 => y=1

\(a,y+\dfrac{2}{3}=\dfrac{5}{2}\)

\(y=\dfrac{5}{2}-\dfrac{2}{3}\)

\(y=\dfrac{15}{6}-\dfrac{4}{6}\)

\(y=\dfrac{11}{6}\)

\(b,3\dfrac{4}{5}-y=\dfrac{18}{5}\)

\(y=3\dfrac{4}{5}-\dfrac{18}{5}\)

\(y=\dfrac{19}{5}-\dfrac{18}{5}\)

\(y=\dfrac{1}{5}\)

\(c,y-4\dfrac{5}{6}=2\dfrac{1}{6}+\dfrac{5}{6}\)

\(y-\dfrac{29}{6}=\dfrac{13}{6}+\dfrac{5}{6}\)

\(y-\dfrac{29}{6}=\dfrac{18}{6}\)

\(y=\dfrac{18}{6}+\dfrac{29}{6}\)

\(y=\dfrac{47}{6}\)

x+y+xy=9

x(y+1)+y+1=10

(x+1)(y+1)=10

x+1,y+1 thuộc Ư(10)

k di rui mik tra loi cho

a)y=5/6

b)y=11/4

\(a\left(\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2}-\frac{1}{10}\right).y=\frac{1}{3}\)

\(\frac{2}{5}.y=\frac{1}{3}\)

      \(y=\frac{1}{3}:\frac{2}{5}\)

     \(y=\frac{5}{6}\)

\(b,\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

     \(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

      \(\frac{10}{11}.y=\frac{2}{3}\)

              \(y=\frac{2}{3}:\frac{10}{11}\)

               \(y=\frac{22}{30}\)