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

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 = |2x-1/3|-7/4

   Do |2x-1/3| \(\ge\) 0

         |2x-1/3|-7/4 \(\ge\)  7/4 

Dấu = xảy ra <=> 2x-1/3=0. =>. x= 1/6

b    1/3|x-2|+2|3-1/2 y|+4

 Do |x-2| \(\ge\) 0

      |3-1/2y| \(\ge\) 0

   => 1/3|x-2|+2|3-1/2 y|+4 \(\ge\) 4

Dấu = xảy ra <=>\(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)

a: Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{6}\)

b: Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)

\(2\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)

Do đó: \(\dfrac{1}{3}\left|x-2\right|+2\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)

\(\Leftrightarrow\left|x-2\right|\cdot\dfrac{1}{3}+\left|3-\dfrac{1}{2}y\right|\cdot2+4\ge4\forall x,y\)

Dấu '=' xảy ra khi x=2 và y=6

=>\(\dfrac{3}{x-5}-\dfrac{y}{3}=\dfrac{1}{6}\)

=>\(\dfrac{9-y\left(x-5\right)}{3\left(x-5\right)}=\dfrac{1}{6}\)

=>9-y(x-5)=1/2(x-5)

=>(x-5)(1/2+y)=9

=>(x-5)(2y+1)=18

=>\(\left(x-5;2y+1\right)\in\left\{\left(18;1\right);\left(-18;-1\right);\left(2;9\right);\left(-2;-9\right);\left(6;3\right);\left(-6;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(23;0\right);\left(-13;-1\right);\left(7;4\right);\left(3;-5\right);\left(11;1\right);\left(-1;-2\right)\right\}\)

\(x^4+2x^3-4x=4\)

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

\(\Leftrightarrow x^2-2=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

\(\Rightarrow x^4+2x^3-4x-4=0\\ \Rightarrow x^4-2x^2+2x^3-4x+2x^2-4=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+2x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=2\\\left(x+1\right)^2+1=0\left(vô.lí\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

=> 1/x = 5/6 - y/3

1/x = 5-2y/6

=> x(5-2y) = 1.6 = 6

Do x ∈ N => x >= 0

Mà 6>0 => 5-2y > 0

Vì y ∈ N => 5-2y ∈ N*

Ta có bảng:

Do x,y ∈ N => (x,y) = (6,2) (thử lại thỏa mãn)

Vậy x=6; y = 2