a) Vì \(\left|x-5\right|\ge0\)nên \(100-\left|x-5\right|\le100\)

Để A lớn nhất thì \(\left|x-5\right|=0\Leftrightarrow x=-5\)

Vậy A lớn nhất bằng 100 khi và chỉ khi x= -5

b) Vì \(\left|y-3\right|\ge0\)nên \(\left|y-3\right|+50\ge50\)

Để B lớn nhất thì \(\left|y-3\right|=0\Leftrightarrow y=3\)

Vậy B nhỏ nhất bằng 50 khi và chỉ khi y= 3

6xy-4x+3y=-53

=>2x(3y-2)+3y-2=-55

=>(3y-2)(2x+1)=-55

=>\(\left(2x+1\right)\left(3y-2\right)=1\cdot\left(-55\right)=\left(-1\right)\cdot55=\left(-55\right)\cdot1=55\cdot\left(-1\right)=5\cdot\left(-11\right)=\left(-11\right)\cdot5=\left(-5\right)\cdot11=11\cdot\left(-5\right)\)

=>\(\left(2x+1;3y-2\right)\in\){(1;-55);(-1;55);(-55;1);(55;-1);(5;-11);(-11;5);(-5;11);(11;-5)}

=>\(\left(x;y\right)\in\left\{\left(0;-\dfrac{53}{3}\right);\left(-1;19\right);\left(-28;1\right);\left(27;\dfrac{1}{3}\right);\left(2;-3\right);\left(-6;\dfrac{7}{3}\right);\left(-3;\dfrac{13}{3}\right);\left(5;-1\right)\right\}\)

mà x,y nguyên

nên \(\left(x;y\right)\in\left\{\left(-1;19\right);\left(-28;1\right);\left(2;-3\right);\left(5;-1\right)\right\}\)

Ta có: \(6xy-8x-3y-2=0\)

\(\Leftrightarrow6xy-3y-8x+4-6=0\)

\(\Leftrightarrow3y\left(2x-1\right)-4\left(2x-1\right)=6\)

\(\Leftrightarrow\left(2x-1\right)\left(3y-4\right)=6\)

\(\Leftrightarrow\left(2x-1\right);\left(3y-4\right)\inƯ\left(6\right)\)

\(\Leftrightarrow\left(2x-1\right)\left(3y-4\right)\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

mà 2x-1 lẻ và \(2x-1\ge-1\) \(\forall x\in N\)

nên \(\left(2x-1\right)\in\left\{-1;1;-3;3\right\}\) và \(\left(3y-4\right)\in\left\{2;-2;6;-6\right\}\)

Trường hợp 1: 

\(\left\{{}\begin{matrix}2x-1=-1\\3y-4=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

Trường hợp 2: 

\(\left\{{}\begin{matrix}2x-1=1\\3y-4=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\3y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{10}{3}\left(loại\right)\end{matrix}\right.\)

Trường hợp 3: 

\(\left\{{}\begin{matrix}2x-1=-3\\3y-4=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-2\\3y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

Trường hợp 4: 

\(\left\{{}\begin{matrix}2x-1=3\\3y-4=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\left(nhận\right)\end{matrix}\right.\)

Vậy: (x,y)=(2;2)

(2x+1)(3y-2)=-11.5

                     =11.-5

2x+1=5

=> x=2

3y-2=(-11)

=> y=(-3)

Con n truong hop khac thi cau tu trinh bay nhung ko ra x,y la Z

Nho tick nha