ta có \(y^2-2y+3=\left(y-1\right)^2+2>=2\) (1)
mặt khác ta có \(x^2+2x+4=\left(x+1\right)^2+3>=3\) => \(\frac{6}{x^2+2x+4}< =\frac{6}{3}=2\) (2)
từ (1) (2) => VT=VP=2<=> \(\hept{\begin{cases}y=1\\x=-1\end{cases}}\)
giai he phuong trinh sau:\(\hept{\begin{cases}\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{cases}}\)
giai he phuong trinh x/x-1 + 2y/y+2 = 3 va 2x/x-1 - y/y+2 = -4
giai he phuong trinh
2x^2+y^2-3xy+3x-2y+1=0
4x^2-y^2+x+4=cbh(2x+y)+cbh(x+4y)
cbh là Căn bậc hai
giai he phuong trinh
\(\hept{\begin{cases}x^2-4\sqrt{3x-2}+10=2y\\y^2-6\sqrt{4y-3}+11=x\end{cases}}\)
\(\hept{\begin{cases}1\le|5x-4|\le3\\\frac{x+1}{2}>\frac{2x+6}{3}\end{cases}}\)GIAI HE BAT PHUONG TRINH
giai he phuong trinh
\(\hept{\begin{cases}\frac{2x-3}{x-2}-\frac{1}{y+2}=7\\\frac{2}{x-2}-\frac{3y+7}{y+2}=13\end{cases}}\)
Giai he phuong trinh
x(x+y)+can (x+y)=can (2y) [can(2y^3)+1]
8x^2-8y+2(x^3-y^3)+3=8y can(2x^3-3x+1)
tat ca deu la can bac hai
giai he phuong trinh
3x2+2y2-4xy+x+8y-4=0
x2-y2+2x+y-3=0
giai he phuong trinh 2x^2-xy=xy^2_2x+y
(x^2+2y^2)(1+1/xy)^2=3
