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\(+)3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)
\(+)7y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{7}\)
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{x}{2}=\dfrac{5y}{15};\dfrac{3y}{15}=\dfrac{z}{7}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y-z}{10+15-21}=\dfrac{32}{16}=2\)
Suy ra: \(\dfrac{x}{10}=2\Rightarrow x=20\)
\(\dfrac{y}{15}=2\Rightarrow y=30\)
\(\dfrac{z}{21}=3\Rightarrow z=63\)
Vậy \(x=20;y=30;z=63\)
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\(a,2x+15=27\)
\(\Rightarrow2x=27-15\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=12\div2\)
\(\Rightarrow x=6\)
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\(xy+12=x+y\)
\(xy-x-y=12\)
\(x\left(y-1\right)-y-1=12-1\)
\(x\left(y-1\right)-\left(y-1\right)=11\)
\(\left(y-1\right)\left(x-1\right)=11\)
Vì \(x,y\in Z\Leftrightarrow y-1;x-1\inƯ\left(11\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y-1=1\\x-1=11\end{matrix}\right.\\\left\{{}\begin{matrix}y-1=-1\\x-1=-11\end{matrix}\right.\\\left\{{}\begin{matrix}y-1=11\\x-1=1\end{matrix}\right.\\\left\{{}\begin{matrix}y-1=-11\\x-1=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2\\x=12\end{matrix}\right.\\\left\{{}\begin{matrix}y=0\\x=-10\end{matrix}\right.\\\left\{{}\begin{matrix}y=2\\x=12\end{matrix}\right.\\\left\{{}\begin{matrix}y=-10\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy ..
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a) 2x + y + 3xy = 5
6x + 3y + 9xy = 15
(9xy + 6x) + 3y = 15
3x(3y+2) + (3y+2) = 17
(3x+2)(3y+1) = 17
bạn tự giải tiếp nhé
a, 2x+y+3xy=5
<=>6x+3y+9xy=3.5
<=>3x(1+3y)+3y+1=15+1
<=>3x(1+3y)+(1+3y)=16
<=>(3x+1)(1+3y)=16
=>3x+1,1+3y thuộc Ư(16)
Vì 3x + 1 chia 3 dư 1 => 3x + 1 thuộc {1;-2;4;-8;16}
=> 1 + 3y thuộc {16;-8;4;-2;1}
Lâp bảng:
3x+1 | 1 | -2 | 4 | -8 | 16 |
1+3y | 16 | -8 | 4 | -2 | 1 |
x | 0 | -1 | 1 | -3 | 5 |
y | 5 | -3 | 1 | -1 | 0 |
Vậy các cặp (x;y) là (0;5);(-1;-3);(1;1);(-3;-1);(5;0)
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Làm 1 câu thôi các câu sau tương tự :v
\(xy+2x+2y=3\)
\(xy+2x+2y+4=7\)
\(x\left(y+2\right)+2\left(y+2\right)=7\)
\(\left(x+2\right)\left(y+2\right)=7\)
\(\Rightarrow x+2;y+2\inƯ\left(7\right)\)
Tìm Ư(7) rồi tính ra nha bạn
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5x + 3x + 6x = 280
14x = 280
x = 280 : 14
x = 20
Vậy x = 20
( 2x - 18 ) - 125 = 325
( 2x - 18 ) = 325 + 125
2x - 18 = 450
2x = 450 + 18
2x = 468
x = 234
Vậy x = 234
\(5x+3x+6x=280\)
\(\Rightarrow x\left(5+3+6\right)=280\)
\(\Rightarrow x.14=280\)
\(\Rightarrow x=20\)
b, \(\left(2x-18\right)-125=325\)
\(\Rightarrow2x-18=450\)
\(\Rightarrow2x=468\)
\(\Rightarrow x=234\)
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\(=3x-2y+5x-y-7y+2x=-10y=-20210\)
\(N=3x-2y+5x-y-7y+2x=7x-10y\\ N=7\cdot2021-10\cdot2021=\left(-3\right)\cdot2021=-6063\)