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ta có :xy-2x+3y=13
xy+3y-2x=13
y(x+3)-2x=13
y(x+3)-2x+6-6=13
y(x+3)-2(x+3)-6=13
(x+3)(y-2)=13+6=19
\(\Rightarrow\left(x+3\right)\left(y-2\right)\inƯ\left(19\right)\)\(=\left(-19;19;1;-1\right)\)
X+3 | 19 | -19 | 1 | -1 |
Y-2 | 1 | -1 | 19 | -19 |
x | 16 | -21 | -2 | -4 |
y | 3 | 1 | 21 | -17 |
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có:
4s=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.........+k(k+1)(k+2)((k+3)-(k-1))
4s=1.2.3.4-1.2.3.0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+........+k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)
4s=k(k+1)(k+2)(k+3)
ta biết rằng tích 4 số tự nhiên liên tiếp khi cộng thêm 1 luôn là 1 số chính phương
=>4s+1 là 1 số chính phương
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x.y+2y+x=6\)
\(\Rightarrow y.\left(x+2\right)+\left(x+2\right)-2=6\)
\(\Rightarrow y.\left(x+2\right)+\left(x+2\right)=8\)
\(\Rightarrow\left(x+2\right).\left(y+1\right)=8\)
\(\Rightarrow\left(x+2\right).\left(y+1\right)\inƯ\left(8\right)=\left\{1;2;4;8\right\}\) mà : \(x+2\ge2\)
\(\Rightarrow\) \(x+2=2\Rightarrow x=0\)
\(y+1=4\Rightarrow y=3\)
\(\Rightarrow x=0;y=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3^{2x+1}+3^{2x}=36\)
\(\Leftrightarrow3^{2x}\left(3+1\right)=36\)
\(\Leftrightarrow3^{2x}.4=36\)
\(\Leftrightarrow3^{2x}=9\)
\(\Leftrightarrow3^{2x}=3^2\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
b) \(\left(x+1\right)^5=\left(x+1\right)^3\)
\(\Leftrightarrow\left(x+1\right)^5-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3\left[\left(x+1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x\in\left\{-2;0\right\}\end{cases}}\)
a)32x + 1 + 32x = 36
=> 32x . 3 + 32x = 36
=> 32x . 4 = 36
=> 32x = 36 : 4
=> 32x = 9
=> 32x = 32
=> 2x = 2
=> x = 1
b) (x + 1)5 = (x + 1)3
=> (x + 1)5 - (x + 1)3 = 0
=> (x + 1)3.[(x + 1)2 - 1] = 0
=>(x + 1)3(x + 1 - 1)(x + 1 + 1) = 0
=> (x + 1)3.x.(x + 2) = 0
=> x = -1 hoặc x = 0 hoặc x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(7x-3xy+3y=19\)
\(\Rightarrow\left(7x-7\right)-\left(3xy-3y\right)=19-7\)
\(7\left(x-1\right)-3y\left(x-1\right)=12\)
\(\left(7-3y\right)\left(x-1\right)=12\)
\(\Rightarrow7-3y;x-1\in\text{Ư}\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Bạn tự lập bảng rồi làm nốt nhé !
xy - 2x - 3y = 1
<=> x(y - 2) - 3y + 6 = 7
<=> x(y - 2) - 3(y - 2) = 7
<=> (y - 2)(x - 3) = 7
Ta có bảng sau :
y - 2
Vậy ...................
xy-2x-3y=1
\(\Rightarrow\)x(y-2)-(3y+1)=0
\(\Rightarrow\)x(y-2)-(3y+6)-5=0
\(\Rightarrow\)x(y-2)-3(y-2)-5=0
\(\Rightarrow\)(x-3)(y-2)=5
\(\Rightarrow\)x-3 và y-2 \(\in\)Ư(5)=5;-5;-1;1
Sau đó bn lập bảng giá trị là OK !!!
Tích nhanh NHA ~~