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Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{2xy}=\frac{1}{2}\)
\(\Rightarrow2y+2x+1=xy\)
\(\Leftrightarrow2y+1=xy-2x\)
\(\Leftrightarrow2y+1=x\left(y-2\right)\)
\(\Leftrightarrow x=\frac{2y+1}{y-2}=\frac{2y-4+5}{y-2}=\frac{2y-4}{y-2}+\frac{5}{y-2}=2+\frac{5}{y-2}\)
Vì x nguyên nên 5 chia hết cho y - 2
Suy ra : y - 2 thuộc Ư(5) = {-5;-1;1;5}
Ta có bảng :
y - 2 | -5 | -1 | 1 | 5 |
y | -3 | 1 | 3 | 7 |
\(x=\frac{2y+1}{y-2}\) | 1 | -3 | 7 | 3 |

Ta có :
\(x+y=\frac{1}{2}\)
\(y+z=\frac{1}{3}\)
\(z+x=\frac{1}{4}\)
\(\Rightarrow\)\(x+y+y+z+z+x=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\Rightarrow\)\(2x+2y+2z=\frac{13}{12}\)
\(\Rightarrow\)\(2\left(x+y+z\right)=\frac{13}{12}\)
\(\Rightarrow\)\(x+y+z=\frac{13}{12}:2\)
\(\Rightarrow\)\(x+y+z=\frac{13}{24}\)
Do đó :
\(x+y+z=\frac{13}{24}\)
\(\Rightarrow\)\(x=\frac{13}{24}-\left(y+z\right)=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\)
\(\Rightarrow\)\(y=\frac{13}{24}-\left(z+x\right)=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\)
\(\Rightarrow\)\(z=\frac{13}{24}-\left(x+y\right)=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\)
Vậy \(x=\frac{5}{24};y=\frac{7}{24};z=\frac{1}{24}\)
Chúc bạn học tốt ~

Bài 1:
\(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{26}\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
......????

d) \(x.\left(y+2\right)-y=15\)
\(\Rightarrow x.\left(y+2\right)=15+y\)
\(\Rightarrow x=\frac{y+15}{y+2}=\frac{y+2+13}{y+2}=1+\frac{13}{y+2}\)
y + 2 là ước nguyên của 13
\(y+2=1\Rightarrow y=-1\Rightarrow x=14\)
\(y+2=-1\Rightarrow y=-3\Rightarrow x=-12\)
\(y+2=13\Rightarrow y=11\Rightarrow x=2\)
\(y+2=-13\Rightarrow y=-15\Rightarrow x=0\)
Ai thấy đúng thì ủng hộ, mink chỉ làm được vậy thuu
\(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)
\(\leftrightarrow\) \(\frac{2}{x}+\frac{2}{y}=1\)
\(\leftrightarrow\)2x + 2y = xy
\(\leftrightarrow\) xy - 2x - 2y + 4 = 4
\(\leftrightarrow\)( x - 2 ) . ( y - 2 ) = 4
\(\leftrightarrow\)x , y ∈ Ư(4)
Ư(4) = { -1 ; -4 ; -2 ; 2 ; 1 ; 4 }
Vậy ta có các cặp x , y thỏa mãn : (2;2),(-2;-2),(1;4), (-1;-4) .
:)
(^^^)