a, 2\(xy\) - 2\(x\) + 3\(y\) = -9

(2\(xy\) - 2\(x\)) + 3\(y\) - 3 = -12

2\(x\)(\(y-1\)) + 3(\(y-1\)) = -12

(\(y-1\))(2\(x\) + 3) = -12

Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12}

Lập bảng ta có:

Theo bảng trên ta có: Các cặp \(x\);\(y\) nguyên thỏa mãn đề bài là:

(\(x;y\)) = (-1; -11); (0; -3); (-3; 5); ( -2; 13)

b, (\(x+1\))²(\(y\) - 3) = -4 

    Ư(4) = {-4; -2; -1; 1; 2; 4}

Lập bảng ta có: 

Theo bảng trên ta có: các cặp \(x;y\) nguyên thỏa mãn đề bài là: 

(\(x;y\)) = (0; -1); (-3; 2); (1; 2)

c) \(\left(x+3\right)^2+\left(2y-1\right)^2< 44\)

\(\Leftrightarrow\left(x+3\right)^2< 44-\left(2y-1\right)^2< 44\) (do \(-\left(2y-1\right)^2\le0\)) (1) 

mà (x + 3)² là số chính phương 

Kết hợp (1) ta được \(\left(x+3\right)^2\le36\)

\(\Leftrightarrow\left(x+3\right)^2\le6^2\Leftrightarrow\left(x+3\right)^2\in\left\{0;1;4;9;25;36\right\}\)

Với (x + 3)² \(\in\left\{0;1;4\right\}\) ta được (2y - 1)² \(\in\left\{0;1;4;9;25;36\right\}\) 

Với (x + 3)² \(\in\left\{9;16\right\}\) ta được (2y - 1)² \(\in\left\{0;1;4;9;25\right\}\) 

Với (x + 3)² = 25 ta được (2y - 1)² \(\in\left\{0;1;4;9;16\right\}\)

Với (x + 3)² = 36 ta được (2y - 1)² \(\in\left\{0;1;4;9\right\}\)

Ta có: 2x + 3y + 5z - 119 = 0

=>  2x + 3y + 5z = 119

 \(\frac{x+2}{3}=\frac{y+3}{5}=\frac{z-4}{7}\Leftrightarrow\frac{2x+4}{6}=\frac{3y+9}{15}=\frac{5z-20}{35}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{2x+4}{6}=\frac{3y+9}{15}=\frac{5z-20}{35}=\frac{2x+4+3y+9+5z-20}{6+15+35}=\frac{119+4+9-20}{56}=\frac{112}{56}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x+2}{3}=2\\\frac{y+3}{5}=2\\\frac{z-4}{7}=2\end{cases}\Rightarrow}\hept{\begin{cases}x+2=6\\y+3=10\\z-4=14\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\y=7\\z=18\end{cases}}\)

Vậy...

1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)

a) \(\left|3x-4\right|+\left|3y+5\right|=0\)

\(\Rightarrow\hept{\begin{cases}3x-4=0\\3y+5=0\end{cases}\Rightarrow\hept{\begin{cases}3x=4\\3y=-5\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{4}{3}\\y=\frac{-5}{3}\end{cases}}\)

b) \(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)

\(\Rightarrow\hept{\begin{cases}x-y=0\\y+\frac{9}{25}=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=\frac{-9}{25}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{-9}{25}\\y=\frac{-9}{25}\end{cases}}}\)

c) \(\left|3-2x\right|+\left|4y+5\right|=0\)

\(\Rightarrow\hept{\begin{cases}3-2x=0\\4y+5=0\end{cases}\Rightarrow\hept{\begin{cases}2x=3\\4y=-5\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{-5}{4}\end{cases}}}\)

d) \(\left|5-\frac{3}{4}x\right|+\left|\frac{2}{7}y-3\right|=0\)

\(\Rightarrow\hept{\begin{cases}5-\frac{3}{4}x=0\\\frac{2}{7}y-3=0\end{cases}\Rightarrow\hept{\begin{cases}\frac{3}{4}x=5\\\frac{2}{7}y=3\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{20}{3}\\y=\frac{21}{2}\end{cases}}\)

e) \(\left(x-1\right)^2+\left(y+3\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=-3\end{cases}}}\)

cảm ơn

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