\(\dfrac{1}{x}-\dfrac{y}{11}=-\dfrac{2}{11}\)

\(\dfrac{1}{x}=-\dfrac{2}{11}+\dfrac{y}{11}\)

\(\dfrac{1}{x}=\dfrac{y-2}{11}\)

\(x\left(y-2\right)=11\)

\(\Rightarrow x,\left(y-2\right)\inƯ\left(11\right)=\left\{1,-1,11,-11\right\}\)

có bảng sau :

Vậy ...

\(\dfrac{1}{x}-\dfrac{y}{11}=-\dfrac{2}{11}\Rightarrow11-xy=-2x\)

\(\Leftrightarrow-2x+xy=11\Leftrightarrow x\left(-2+y\right)=11\)

\(\Rightarrow x;y-2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

|x(x-4)|=x

=> x(x-4)=x               hoặc x(x-4)=-x

=> x²-4x-x=0             hoặc x²-4x+x=0

=> x²-5x=0                hoặc x²-3x=0

=> x(x-5)=0                hoặc x(x-3)=0

=> x=0 hay x-5=0      hoặc x=0  hay x-3=0

=> x=0 hay x=0+5     hoặc x=0  hayc x=0+3

=> x=0 hay x=5         hoặc x=0 hay x=3 

=> x \(\in\){0;3;5}

\x(x-4)\=x<=>x-4=1 hoac=-1

xet :x-4=1=> x=-3(vli)

      :x-4=-1=>x=3

=> x=3

\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)

\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)

7\(x\) < 36 < 63\(x\) + 7

⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)

\(\dfrac{29}{63}\)<  \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}

⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)

\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)

=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)

\(\Rightarrow7x< 36< 7x+7\)

\(\Rightarrow x< \dfrac{36}{7}< x+1\)

\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)

\(\Rightarrow x=5\)

tik cho mình nhé

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\(VT=\sqrt{14}-\sqrt{13}=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)

\(VP=2\sqrt{3}-\sqrt{11}=\sqrt{12}-\sqrt{11}=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)

Ta thấy: \(\sqrt{14}+\sqrt{13}>\sqrt{12}+\sqrt{11}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{14}+\sqrt{13}}< \dfrac{1}{\sqrt{12}+\sqrt{11}}\)

Hay \(VT< VP\)

Vậy \(\sqrt{14}-\sqrt{13}< 2\sqrt{3}-\sqrt{11}\)

\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\\ \Leftrightarrow6x-42=7y-42\\ \Leftrightarrow6x=7y\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{7}=\dfrac{y}{6}=\dfrac{x-y}{7-6}=\dfrac{-4}{1}=-4\\ \dfrac{x}{7}=-4\Leftrightarrow x=-28\\ \dfrac{y}{6}=-4\Leftrightarrow y=-24\)

a  tìm số nguyên x biết (x-5).(y-7)=1 
   (x-5).(y-7)=1 = 1.1 = -1.(-1) 
   TH1,
   x-5 = 1, y-7 = 1
   => x = 6, y = 8
   TH2