xy = 3x + 5y = 2
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4.2-3.2+5.0=2
mình ko bt đúng hay sai nhưng mình tính ra thế đó
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x-5y\right)^2+\left(xy-135\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(3x-5y\right)^2\ge0\forall x,y\\\left(xy-135\right)^2\ge0\forall x,y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3x-5y=0\\xy-135=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\xy=135\end{matrix}\right.\)
\(\Rightarrow\dfrac{5}{3}y.y=135\)\(\Rightarrow y^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}y=9\Rightarrow x=15\\y=-9\Rightarrow x=-15\end{matrix}\right.\)
Ta có: \(\left(3x-5y\right)^2\ge0\forall x;y\)
\(\left(xy-135\right)^2\ge0\forall x;y\)
\(\Rightarrow\left(3x-5y\right)^2+\left(xy-135\right)^2\ge0\forall x\)
Mặt khác: \(\left(3x-5y\right)^2+\left(xy-135\right) ^2=0\)
nên: \(\left\{{}\begin{matrix}\left(3x-5y\right)^2=0\\\left(xy-135\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-5y=0\\xy-135=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=5y\\xy=135\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\\dfrac{5}{3}y^2=135\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\y^2=81\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\end{matrix}\right.\)
\(+,TH1:y=9\Leftrightarrow x=\dfrac{5}{3}\cdot9=15\left(tm\right)\)
\(+,TH2:y=-9\Leftrightarrow x=\dfrac{5}{3}\cdot\left(-9\right)=-15\left(tm\right)\)
Vậy ...
#\(Toru\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình chỉ phân tích hộ bạn, rồi bạn tự lập bảng và tìm ra giá trị x;y nhé :)
a) xy + x + y = 2
<=> xy + x + y + 1 = 2
<=> x ( y + 1 ) + ( y + 1 ) = 2
<=> ( x + 1 )( y + 1) = 2
b) xy - 10 + 5x - 3y = 2
<=> xy - 3y + 5x - 15 = -3
<=> y ( x - 3 ) + 5 ( x - 3 ) = -3
<=> ( x - 3 )( y + 5 ) = -3
c) xy - 1 = 3x + 5y + 4
<=> xy - 3x - 5y = 5
<=> xy - 3x - 5y + 15 = -10
<=> x ( y - 3 ) - 5 ( y - 3 ) = -10
<=> ( x - 5 ) ( y - 3 ) = -10
d) 3x + 4y - xy = 15
<=> 3x - xy - 12 + 4y = 3
<=> x ( 3 -y ) - 4 ( 3 - y ) = 3
<=> ( x - 4 ) ( 3 - y ) = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(xy+3x=5y-2\)
\(\Leftrightarrow x\left(y+3\right)-5y+2=0\)
\(\Leftrightarrow x\left(y+3\right)-5y-15+17=0\)
\(\Leftrightarrow\left(x-5\right)\left(y+3\right)=-17\)
Lại có:-17 = (-1) . 17 = 17 . (-1) = (-17) . 1 = 1 . (-17)
Thay vào là tìm đc
![](https://rs.olm.vn/images/avt/0.png?1311)
xy+x+y=2
xy+x+y+1=2+1
(xy+x)+(y+1)=3
x(y+1)+(y+1)=3
(x+1)(y+1)=3=1.3=3.1=-1.-3=-3.-1
\(\Rightarrow\left[{}\begin{matrix}x+1=1;y+1=3\\x+1=3;y+1=1\\x+1=-1;y+1=-3\\x+1=-3;y+1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0;y=2\\x=2;y=0\\x=-2;y=-4\\x=-4;y=-2\end{matrix}\right.\)
Vậy:.................
xy+14+2y+7x= -10
\(\Leftrightarrow\)y(x+2)+7(x+2)=-10
\(\Leftrightarrow\)(y+7)(x+2)=-10=1.(-10)=2.(-5)=5.(-2)=10.(-1)
y+7 | 1 | 2 | 5 | 10 |
x+2 | -10 | -5 | -2 | -1 |
y | -6 | -5 | -2 | 3 |
x | -12 | -7 | -4 | -3 |
![](https://rs.olm.vn/images/avt/0.png?1311)
a: (3x^2-4)(x+3y)
=3x^2*x+3x^2*3y-4x-4*3y
=3x^3+9x^2y-4x-12y
b: (c+3)(x^2+3x)
=c*x^2+c*3x+3x^2+9x
=cx^2+3cx+3x^2+9x
c: (xy-1)(xy+5)
=xy*xy+5xy-xy-5
=x^2y^2+4xy-5
d: (3x+5y)(2x-7y)
=3x*2x-3x*7y+5y*2x-5y*7y
=6x^2-21xy+10xy-35y^2
=6x^2-11xy-35y^2
e: -(x-1)(-x^2+2y)
=(x-1)(x^2-2y)
=x^3-2xy-x^2+2y
f: (-x^2+2y)(x^2+2y)
=(2y)^2-x^4
=4y^2-x^4
![](https://rs.olm.vn/images/avt/0.png?1311)
=\(^{\dfrac{-x^2-xy}{5\left(x^2-y^2\right)}}\).\(\dfrac{3\left(x^3-y^3\right)}{x^2-xy}\)
=\(\dfrac{-3\left(x-y\right)}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)