\(a\orbr{x=\frac{\pm\sqrt{5}-3}{4}}\)

\(b\hept{\begin{cases}x=5\\y=4\end{cases}}\)

2)\(\Leftrightarrow\left(x^3-x^2y\right)+\left(y^3-xy^2\right)=5\)

\(\Leftrightarrow x^2\left(x-y\right)+y^2\left(y-x\right)=5\)

\(\Leftrightarrow x^2\left(x-y\right)-y^2\left(x-y\right)=5\)

\(\Leftrightarrow\left(x-y\right)\left(x^2-y^2\right)=5\)

TH1\(\hept{\begin{cases}x-y=1\\x^2-y^2=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}\left(N\right)}}\)

TH2\(\hept{\begin{cases}x-y=5\\x^2-y^2=1\end{cases}\Leftrightarrow\hept{ }x,y\in\varnothing}\)

TH3\(\hept{\begin{cases}x-y=-1\\x^2-y^2=-5\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}\left(N\right)}}\)

TH4\(\hept{\begin{cases}x-y=-5\\x^2-y^2=-1\end{cases}\Leftrightarrow\hept{ }x,y\in\varnothing}\)

Vậy......

bạn mai anh làm đúng rồi mình xét thiếu trường hợp . nhưng nên phân tích thành (x+y)(x-y)² dễ hơn

đặt x+y=a

xy=b

ntc a-2

chụp cho tớ 20 bài bđt đi chi

Đặt x=y=-2, pt trở thành: 

\(\left(x+2\right)^2z+\left(z+2\right)^2x+26=0\Leftrightarrow\left(x+z+8\right)\left(xz+4\right)=6\)\(\Rightarrow x+z+8\in U\left(6\right)\)

Giải các TH ta thu được cặp số (x;y) thoả mãn đk là:

(x;y)=(1;-1), (3,-3), (-10;3), (1;-8)

Bài 1:

Đặt 2x+1=a

Theo đề, ta có: \(\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}=3\)

=>3a^2(a+1)^2=a^2+2a+1+a^2

=>3a^2(a^2+2a+1)-2a^2-2a-1=0

=>3a^4+6a^3+a^2-2a-1=0

=>(a^2+a-1)(3a^2+3a+1)=0

=>\(a\in\left\{\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)

=>\(2x+1\in\left\{\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)

=>\(2x\in\left\{\dfrac{-3+\sqrt{5}}{2};\dfrac{-3-\sqrt{5}}{2}\right\}\)

hay \(x\in\left\{\dfrac{-3+\sqrt{5}}{4};\dfrac{-3-\sqrt{5}}{4}\right\}\)

\(PT\Leftrightarrow x^3+2x^2+3x+2=y^3\)

Với  x thuộc đoạn {-1,1} ta có

\(x^3< x^3+2x^2+3x+2< \left(x+1\right)^3\)

\(\Rightarrow x^3< y^3< \left(x+1\right)^3\)(vô lí)

\(\Rightarrow x\in[-1;1]\)

\(\Rightarrow x\in\left\{-1,0,1\right\}\)

Với x=-1=> y=0(tm)

Với x=0=>\(y=\sqrt[3]{2}\left(ktm\right)\)

Với x=1=>y=2(tm)

Vậy...........

\(\Leftrightarrow\left\{{}\begin{matrix}xy-3x+2y-6=xy+1\\2x+2y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y-3x=7\\2x+2y=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{29}{10}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x+2\right)\left(y-3\right)=xy+1\\2\left(x+y\right)\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y-3\right)=xy+1\\2x+2y=5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}xy-3x+2y-6=xy+1\\y=\dfrac{5-2x}{2}\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}-3x+2y=7\\y=\dfrac{5-2x}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-3x+2.\dfrac{5-2x}{2}=7\\y=\dfrac{5-2x}{2}\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=-0,4\\y=2,9\end{matrix}\right.\)