3x + y = 16 (1)
2x - y = -1 (2)
(1) ⇔ y = 16 - 3x (3)
Thế (3) vào (2), ta có:
2x - (16 - 3x) = -1
⇔ 2x - 16 + 3x = -1
⇔ 5x = -1 + 16
⇔ 5x = 15
⇔ x = 15 : 5
⇔ x = 3
Thế x = 3 vào (3), ta có:
y = 16 - 3.3
⇔ y = 7
Vậy S = {(3; 7)}
\(\left\{{}\begin{matrix}x^3+xy^2+x^2+3x=2y^3+2x^2y+6y\\2\sqrt{y-1}+6\sqrt{xy-5x+3}=x^2+12x-16\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3+xy^2+x^2+3x=2y^3+2x^2y+6y\\2\sqrt{y-1}+6\sqrt{xy-5x+3}=x^2+12x-16\end{matrix}\right.\)
Cho \(x,y,z>0\)và \(xy+yz+zx=xyz.Cm\)\(\frac{1}{x+2y+3z}+\frac{1}{2x+3y+z}+\frac{1}{3x+y+2z}< \frac{3}{16}\)
giải hệ phương trình
1)\(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\) 2)\(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\) 3)\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}-x+3y=16\\2x+y=3\end{matrix}\right.\) 5)\(\left\{{}\begin{matrix}\dfrac{-3}{x-y}+\dfrac{5}{2x+y}=-2\\\dfrac{4}{x-y}-\dfrac{10}{2x+y}=2\end{matrix}\right.\) 6)\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
giải hpt
\(\hept{\begin{cases}x^2+y^2+\frac{8xy}{x+y}=16\\2x^2-5x+2\sqrt{x+y}-\sqrt{3x-2}=0\end{cases}}\)
tìm x, y biết rằng 2x2 + 2y2 +3x +3y -5xy +16 = 0.
b) Giải hệ phương trình 1/(3x) + (2x)/(3y) = (x + sqrt(y))/(2x ^ 2 + y); 2(2x + sqrt(y)) = sqrt(2x + 6) - y
a) tìm các số nguyên x,t thỏa mãn 2y(2x2-1) - 2x(2y2-1)+1=x3y3
b) giải pt 2x2 +2x+1=(2x+3)(\(\sqrt{x^2+x+2}\)- 1)
c) giải hệ pt \(\hept{\begin{cases}x^2+y^2+\frac{8xy}{x+y}=16\\\sqrt{x^2+12}+\frac{5}{2}\sqrt{x+y}=3x+\sqrt{x^2+5}\end{cases}}\)
Giai hptr : \(\left\{{}\begin{matrix}\left(2y-3\right)\left(3x-4\right)=\left(3y+1\right)\left(2x-5\right)\\2\left(y-3\right)+16=3\left(x+2\right)\end{matrix}\right.\)
2 x + y + 1 x + y = 3 1 x + y - 3 x - y = 1
