Ta có : (2c+1)(3y−2)=5(2c+1)(3y−2)=5

Do c,y∈Z⇒2c+1,3y−2∈Zc,y∈Z⇒2c+1,3y−2∈Z

Mà : 5=1.5=(−1).(−5)5=1.5=(−1).(−5) Nên ta xét các trường hợp :

TH1 : 2c+1=1,3y−2=5⇒c=0,y=732c+1=1,3y−2=5⇒c=0,y=73 ( Loại do c,y∈Zc,y∈Z )

TH2 : 2c+1=5,3y−2=1⇒c=2,y=12c+1=5,3y−2=1⇒c=2,y=1 ( Thỏa mãn )

TH3 : 2c+1=−1,3y−2=−5⇒c=−1,y=−12c+1=−1,3y−2=−5⇒c=−1,y=−1 ( Thỏa mãn )

TH4 : 2c+1=−5,3y−2=−1⇒c=−3,y=132c+1=−5,3y−2=−1⇒c=−3,y=13 ( Loại do c,y∈Zc,y∈Z ) 

Vậy : (c,y)∈(c,y)∈{(2,1);(−1,−1)(2,1);(−1,−1)}

tìm c,y là số nguyên biết 

\(\Leftrightarrow\text{(2c+1).(3y-2)}=0\)

\(\Leftrightarrow\orbr{\begin{cases}2c+1=0\\3x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2c=-1\\3x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)

Vậy...

Học tốt

a) \(3^{x+2}\cdot5^{y-3}=45^x\)

\(\Rightarrow3^{x+2}\cdot5^{y-3}=\left(3^2\right)^x\cdot5^x\)

\(\Rightarrow3^{x+2}\cdot5^{y-3}=3^{2x}\cdot5^x\)

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)

(2x+1) . (3y -2)=-5

=> 2x+1 \(\in\)Ư(-5) = { 1; 5; -1; -5}

=> 2x \(\in\){ 0; 6; -2; -6}

=> x \(\in\){ 0; 3; -1; -3}

Sau bn tự thay nha

\(\left(2x+1\right)\left(3y-2\right)=5\)

Do x,y nguyên => 2x+1; 3y-2 nguyên

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

Vậy (x;y)=(-1;-1);(2;1)

\(a,\left\{{}\begin{matrix}\left|x-3y\right|\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y=-12\\y=-4\end{matrix}\right.\)

\(b,Sửa:\left|x-y-5\right|+\left(y+3\right)^2=0\\ \left\{{}\begin{matrix}\left|x-y-5\right|\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y-5=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=2\\y=-3\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}\left|x+y-1\right|\ge0\\\left(y-2\right)^4\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y=-1\\y=2\end{matrix}\right.\)

\(d,\left\{{}\begin{matrix}\left|x+3y-1\right|\ge0\\3\left|y+2\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+3y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-3y=7\\y=-2\end{matrix}\right.\)

\(e,Sửa:\left|2021-x\right|+\left|2y-2022\right|=0\\ \left\{{}\begin{matrix}\left|2021-x\right|\ge0\\\left|2y-2022\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2021-x=0\\2y-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\y=1011\end{matrix}\right.\)