Lời giải:

Ta thấy:

$10x\equiv 0\pmod 5$

$288\equiv 3\pmod 5$

$\Rightarrow y^2\equiv 3\pmod 5$ (vô lý)

Do ta biết rằng một số chính phương khi chia cho $5$ chỉ có thể có dư là $0,1,4$.

Như vậy, không tồn tại số tự nhiên $x,y$ thỏa mãn điều kiện đề bài.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

Ta có: \(x^2y^2z^2=288^2\)

\(\Leftrightarrow36k^2=288^2\)

\(\Leftrightarrow k^2=2304\)

Trường hợp 1: k=48

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=96\\y=3k=144\\z=6k=288\end{matrix}\right.\)

Trường hợp 2: k=-48

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-96\\y=3k=-144\\z=6k=-288\end{matrix}\right.\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

\(\Rightarrow4k^29k^236k^2=288^2\)

\(\Rightarrow k^6=64\)

\(\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4,y=6,z=12\\x=-4,y=-6,z=-12\end{matrix}\right.\)

a.

Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)

Thế vào \(2x+y-z=81\)

\(\Rightarrow2.5k+3k-4k=81\)

\(\Rightarrow9k=81\)

\(\Rightarrow k=9\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)

b.

Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)

Thế vào \(5x-y+3z=124\)

\(\Rightarrow5.3k-5k+3.2k=124\)

\(\Rightarrow16k=124\)

\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)

c.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Thế vào \(xyz=810\)

\(\Rightarrow2k.3k.5k=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)

d.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

Thế vào \(x^2y^2z^2=288^2\)

\(\Rightarrow\left(2k\right)^2.\left(3k\right)^2.\left(6k\right)^2=288^2\)

\(\Rightarrow\left(k^2\right)^3=64\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=\pm2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=6k=12\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=6k=-12\end{matrix}\right.\)

\(a)\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+x+x+z+2+x+y-3}{x+y+z}\)

\(=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)

Lại có: \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)

\(\Rightarrow2=\dfrac{1}{x+y+z}\Rightarrow2\left(x+y+z\right)=1\Rightarrow x+y+z=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y+z+x+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1+\dfrac{1}{2}}{3}\\y=\dfrac{\dfrac{1}{2}+2}{3}\\z=\dfrac{\dfrac{1}{2}-3}{3}\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)

Chúc bạn học tốt!

\(\dfrac{x}{7}+\dfrac{y}{41}+\dfrac{z}{49}=\dfrac{1000}{2009}\)

\(\Leftrightarrow\dfrac{287x+49y+41z}{2009}=\dfrac{1000}{2009}\)

\(\Leftrightarrow287x+49y+41z=1000\)

\(\Leftrightarrow41z=1000-287x-49y\le1000-287-49=664\) do \(x,y\) nguyên dương. (1)

Mặt khác ta cũng có \(1000\equiv6\left(mod7\right);287\equiv0\left(mod7\right);49\equiv0\left(mod7\right)\)

\(\Rightarrow1000-287x-49y\equiv6\left(mod7\right)\)

Mà \(41\equiv6\left(mod7\right)\Rightarrow z\equiv1\left(mod7\right)\) (2)

Từ (1) suy ra \(1\le z\le\dfrac{664}{41}\le16\) (3)

Từ (2),(3) suy ra \(z\in\left\{8;15\right\}\)

+) \(z=8\Leftrightarrow287x+49y=672\)

\(\Leftrightarrow41x+7y=96\)

Bằng phép thử ta nhận nghiệm \(\left(x;y\right)=\left(2;2\right)\)

+) \(z=15\Leftrightarrow287x+49y=385\)

\(\Leftrightarrow41x+7y=55\)

Bằng phép thử ta nhận nghiệm \(\left(x;y\right)=\left(1;2\right)\)

Vậy tập nghiệm nguyên dương của phương trình là \(\left(x;y;z\right)\in\left\{\left(2;2;8\right);\left(1;2;15\right)\right\}\)

5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{2\cdot2+3\cdot3-4}=\dfrac{45}{9}=5\)

Do đó: x-1=10; y-2=15; z-3=20

=>x=11; y=17; z=23

c: Ta có: 10x=6y

nên x/3=y/5

Đặt x/3=y/5=k

=>x=3k; y=5k

Ta có: \(2x^2-y^2=-28\)

\(\Leftrightarrow2\cdot9k^2-25k^2=-28\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

=>x=6; y=10

TRường hợp 2: k=-2

=>x=-6; y=-10