\(\left\{{}\begin{matrix}x+y+z=1\\x^2+y^2+z^2=1\\x^3+y^3+z^3=1\end{matrix}\right.\)

☘ Ta có:

\(yz=\dfrac{\left(y+z\right)^2-\left(y^2+z^2\right)}{2}\)

\(=\dfrac{\left(1-x\right)^2-\left(1-x^2\right)}{2}=x^2-x\)

☘ Thay vào phương trình thứ 3

\(\Rightarrow1=x^3+y^3+z^3=x^3+\left(y+z\right)^3-3yz\left(y+z\right)\)

\(=x^3+\left(1-x\right)^3-3\left(x^2-x\right)\left(1-x\right)\)

\(=1+3x^3-3x^2\)

\(\Rightarrow3x^2\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

⚠ Chia thành hai trường hợp, rồi tự giải tiếp nhé.

⚠ Nguồn: Ý tưởng xuất phát từ [Báo TTT - số 71 mục "Thi giải toán qua thư"]

⚠ Có thể có cách khác ngắn gọn, dễ hiểu hơn.

✿ Another way ✿

☘ Ta có:

\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)

\(\Rightarrow xy+z\left(x+y\right)=0\)

\(\Rightarrow xy=-z\left(x+y\right)=-z\left(1-z\right)=z^2-z\left(1\right)\)

☘ Mặt khác

\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

\(\Rightarrow xyz=0\left(2\right)\)

☘ Thay (1) vào (2)

\(\Rightarrow z\left(z^2-z\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}z=0\\z=1\end{matrix}\right.\)

⚠ Cũng chia thành hai trường hợp rồi giải tiếp nhé.

sorry , mik chỉ học lớp 6 thôi

Ta có: \(\left\{{}\begin{matrix}x^3+y^2=2\\x^2+y^3=2\end{matrix}\right.\) \(\Rightarrow x^3+y^2=x^2+y^3\Leftrightarrow x^3-x^2=y^3-y^2\Leftrightarrow x^2\left(x-1\right)=y^2\left(y-1\right)\)

\(\Rightarrow\) x=y

\(\Rightarrow\)\(x^3+y^2=2\Leftrightarrow x^3+x^2=2\Leftrightarrow x=1\)\(\Rightarrow y=1\)

\(\left\{{}\begin{matrix}\sqrt{x^2+3}+2\sqrt{x}=3+\sqrt{y}\left(1\right)\\\sqrt{y^2+3}+2\sqrt{y}=3+\sqrt{x}\left(2\right)\end{matrix}\right.\)\(\left(đk;x;y\ge0\right)\)

\(\left(1\right)-\left(2\right)\Rightarrow\sqrt{x^2+3}+2\sqrt{x}-\sqrt{y^2+3}-2\sqrt{y}=\sqrt{y}-\sqrt{x}\)

\(\Leftrightarrow\sqrt{x^2+3}-\sqrt{y^2+3}+2\sqrt{x}-2\sqrt{y}+\sqrt{x}-\sqrt{y}=0\left(3\right)\)

\(với:x=y=0\Rightarrow ko\) \(là\) \(nghiệm\)

\(vỡi:x=y\ne0\Rightarrow x;y>0\)

\(\left(3\right)\Leftrightarrow\dfrac{x^2+3-y^2-3}{\sqrt{x^2+3}+\sqrt{y^2+3}}+\dfrac{4x-4y}{2\sqrt{x}+2\sqrt{y}}+\dfrac{x-y}{\sqrt{x}+\sqrt{y}}=0\)

\(\Leftrightarrow\left(x-y\right)\left[\dfrac{x+y}{\sqrt{x^2+3}+\sqrt{y^2+3}}+\dfrac{4}{2\sqrt{x}+2\sqrt{y}}+\dfrac{1}{\sqrt{x}+\sqrt{y}}>0\left(\forall x;y>0\right)\right]=0\)

\(\Rightarrow x=y\left(4\right)\)

\(\left(4\right)và\left(1\right)\Rightarrow\sqrt{x^2+3}+2\sqrt{x}=3+\sqrt{x}\Leftrightarrow\sqrt{x^2+3}+\sqrt{x}-3=0\)

\(\Leftrightarrow\sqrt{x^2+3}-2+\sqrt{x}-1=0\Leftrightarrow\dfrac{x^2+3-4}{\sqrt{x^2+3}+2}+\dfrac{x-1}{\sqrt{x}+1}=0\Leftrightarrow\left(x-1\right)\left[\dfrac{x+1}{\sqrt{x^2+3}+2}+\dfrac{1}{\sqrt{x}+1}>0\left(\forall x>1\right)\right]=0\Leftrightarrow x=y=1\)

a: x-2y=5 và 3x+y=8

=>3x-6y=15 và 3x+y=8

=>-7y=7 và x-2y=5

=>y=-1 và x=5+2y=5-2=3

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{6}{y-2}=9\\\dfrac{3}{x+1}-\dfrac{1}{y-2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y-2}=7\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=3\end{matrix}\right.\)

=>y-2=1 và x+1=1

=>x=0 và y=3

\(x^4+y^4=1\Rightarrow\left\{{}\begin{matrix}\left|x\right|\le1\\\left|y\right|\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^3\le x^2\\y^3\le y^2\end{matrix}\right.\)

\(\Rightarrow x^3+y^3\le x^2+y^2\)

Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x^4+y^4=1\\x^3=x^2\\y^3=y^2\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)

- Trừ hai pt ta được :\(x^3-y^3-x^2+y^2+x-y+1-1=2y-2x\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x-y\right)\left(x+y\right)+\left(x-y\right)+2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-\left(x+y\right)+3\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-x-y+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x^2+xy+y^2-x-y+3=0\end{matrix}\right.\)

TH1 : x = y

PT ( I ) TT : \(x^3-x^2+x+1-2x=x^3-x^2-x+1=0\)

\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow x=y=\pm1\)

TH₂ : \(x^2+xy+y^2-x-y+3=0\)

\(\Leftrightarrow x^2+\dfrac{y^2}{4}+\dfrac{1}{4}+xy-x-\dfrac{1}{2}y+\dfrac{3}{4}y^2-\dfrac{1}{2}y+\dfrac{11}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2+\dfrac{8}{3}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2=-\dfrac{8}{3}\left(VL\right)\)

Vậy ....

Ta có hệ \(\hept{\begin{cases}x^2+y^2-3x+4y=1\\3x^2-2y^2-9x-8y=3\end{cases}\Leftrightarrow\hept{\begin{cases}3x^2+3y^2-9x+12y=3\left(1\right)\\3x^2-2y^2-9x-8y=3\left(2\right)\end{cases}}}\)

Lấy (1)-(2) ta có \(5y^2+20y=0\Leftrightarrow\orbr{\begin{cases}y=0\\y=-4\end{cases}}\)

Với \(y=0\Rightarrow x^2-3x-1=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}}\)

Với \(y=-4\Rightarrow x^2-3x-1=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}}\)

Vậy hệ có 4 nghiệm \(\left(x;y\right)=\left(0;\frac{3+\sqrt{13}}{2}\right);\left(0;\frac{3-\sqrt{13}}{2}\right);\left(-4;\frac{3+\sqrt{13}}{2}\right);\left(-4;\frac{3-\sqrt{13}}{2}\right)\)