\(\left(2x-1\right)\left(y-7\right)=22\)

\(\Rightarrow\left(2x-1\right);\left(y-7\right)\in\left\{1;2;11;22\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;29\right);\left(\dfrac{3}{2};18\right);\left(6;9\right);\left(\dfrac{23}{2};8\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;29\right);\left(6;9\right)\right\}\left(x;y\inℤ^+\right)\)

cách kiếm xu kiểu gì

bạn lên phần hỏi đáp mà hỏi

a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

\(A=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

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\(x+\frac{1}{y+\frac{1}{z}}=\frac{10}{7}\Leftrightarrow x+\frac{1}{y+\frac{1}{z}}=1+\frac{3}{7}\Leftrightarrow x+\frac{1}{y+\frac{1}{z}}=1+\frac{1}{\frac{7}{3}}\)

\(\Leftrightarrow x+\frac{1}{y+\frac{1}{z}}=1+\frac{1}{2+\frac{1}{3}}\Leftrightarrow x=1;y=2;z=3\)

Tu de bai suy ra 2y+2x=xy<=>...<=>y(2-x)= -2x<=>y=2x/(x-2)<=>y=(2x-4+4)/(x-2)<=>y=2+4/(x-2)

vi x la so nguyen Dưỡng nen x-2 la so nguyen  duong va la ước cua 4 => x-2 =1 hoặc x-2= 4 => x=3 hoac x=6 

Voi x=3 => y= 6

voi x=6=> y=3

vay cac cap so nguyen duong (x;y) can tim la (3;6); (6;3)

.....

Sau khi chi ra x-2 la uoc nguyen duong cua 4

 Co 3  Truong hop

x-2 =1; x-2=2;x-2=4

Tu do tinh duoc x=3;x=4;x=6. Suy ra cac gia tri tuong ung cua y

co 3 cap so nguyen duong x, y can Tim:(3;6);(4 ;4);(6;3)