Ai giúp bài câu này với (xy+5).(-1+xy)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=> xy-x+2y-2=5-2(trừ cả 2 vế cho 2 bn nhé)
=> y(x+2)-(x+2)=3
=> (x+2)(y-1)=3
=>\(\hept{\begin{cases}x+2=1\\y-1=3\end{cases}}\)=>\(\orbr{\begin{cases}x=-2\\y=4\end{cases}}\)
\(\hept{\begin{cases}x+2=3\\y-1=1\end{cases}}\)=>\(\hept{\begin{cases}x=1\\y=2\end{cases}}\)
\(\orbr{\begin{cases}x+2=-1\\y-1=-3\end{cases}}\)=> \(\hept{\begin{cases}x=-3\\y=-2\end{cases}}\)
\(\hept{\begin{cases}x+2=-3\\y-1=-1\end{cases}}\)=>\(\hept{\begin{cases}x=-5\\y=0\end{cases}}\)
\(\left(xy-x\right)+\left(2y-2\right)=3\)
\(x.\left(y-1\right)+2.\left(y-1\right)=3\)
\(\left(y-1\right).\left(x+2\right)=3\)
\(\left(y-1\right).\left(x+2\right)=3.1=1.3=-3.-1=-1.-3\)
x+2 | 3 | 1 | -1 | -3 |
x | 1 | -1 | -3 | -5 |
y-1 | 1 | 3 | -3 | -1 |
y | 2 | 4 | -2 | 0 |
vậy các cặp (x;y) phải tìm lag (1;2) ; (-1;4) ; (-3;-2) ; (-5;0)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(6x^2+xy-7x-2y^2+7y-5=-\left(y-2x-1\right)\left(2y+3x-5\right)\)
\(6x^2+xy-7x-2y^2+7y-5=-2y\left(y-2x-1\right)-3x\left(y-2x-1\right)+5\left(y-2x-1\right)=-\left(y-2x-1\right)\left(2y+3x-5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(xy+x+y=4\\ x\left(y+1\right)+y+1=4+1=5\\ \left(x+1\right)\left(y+1\right)=5\)
\(x+1\) | \(5\) | \(1\) | \(-1\) | \(-5\) |
\(y+1\) | \(1\) | \(5\) | \(-5\) | \(-1\) |
\(x\) | \(4\) | \(0\) | \(-2\) | \(-6\) |
\(y\) | \(0\) | \(4\) | \(-6\) | \(-2\) |
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn cho mik hỏi: bài này x,y thuộc N hay x,y thuộc Z vậy ?
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}+1\right)\)
\(:\left(1-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}+\dfrac{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right)\)
\(:\left(\dfrac{\text{}\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right)\)
\(.\left(\dfrac{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}\right)\)
\(A=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(xy+x+y=5\)
\(\Rightarrow x\left(y+1\right)+y=5\)
\(\Rightarrow x\left(y+1\right)+\left(y+1\right)=6\)
\(\Rightarrow\left(x+1\right).\left(y+1\right)=6\)
x+1 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
y+1 | -1 | -2 | -3 | -6 | 6 | 3 | 2 | 1 |
x | -7 | -4 | -3 | -2 | 0 | 1 | 2 | 5 |
y | -2 | -3 | -4 | -7 | 5 | 2 | 1 | 0 |
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
xy + yx = 10x + y + 10y + x = 11x + 11 y \(⋮\)11
Vậy xy + yx \(⋮11\)
( đpcm )
Có : A = 10.x+y+10.y+y=11x+11y=11.(x+y) chia hết cho 11
=>ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
có \(\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}\)
=>\(\hept{\begin{cases}x⋮y\\y⋮x\end{cases}}\)=>x=y
Thay y=x vào A:\(\frac{x^2+2019x^2}{x\cdot x}=\frac{2020\cdot x^2}{x^2}=2020\)
Vậy A=2020
Bạn cần làm gì với phép tính này thì bạn cần ghi rõ ra.