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

☘ 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.

Câu 1:

\(ĐK:x\ge2\)

Áp dụng BĐT cauchy ta có:

\(\left(x+1\right)+4\ge2\sqrt{4\left(x+1\right)}=4\sqrt{x+1}\\ \Leftrightarrow2\sqrt{x+1}\le\dfrac{x+5}{2}\)

Ta có \(\left(x-2\right)+1\ge2\sqrt{x-2}\Leftrightarrow\sqrt{x-2}\le\dfrac{x-1}{2}\)

\(\Leftrightarrow P\le\dfrac{x+5}{2}+\dfrac{x-1}{2}-x+2013=x+2-x+2013=2015\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x-2=1\end{matrix}\right.\Leftrightarrow x=3\)

Câu 2:

\(HPT\Leftrightarrow\left\{{}\begin{matrix}10\sqrt{x}+15y^3=140\\4y^3-10\sqrt{x}=12\end{matrix}\right.\left(x\ge0\right)\\ \Leftrightarrow19y^3=152\\ \Leftrightarrow y^3=8\Leftrightarrow y=2\\ \Leftrightarrow2\sqrt{x}+24=28\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

Vậy \(\left(x;y\right)=\left(4;2\right)\)

Câu 3:

\(HPT\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\my+2m+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=\dfrac{3-2m}{m+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{m+1}\\x=\dfrac{3-2m}{m+1}\end{matrix}\right.\\ \Leftrightarrow xy=\dfrac{5\left(3-2m\right)}{\left(m+1\right)^2}\)

Đặt \(xy=t\)

\(\Leftrightarrow m^2t+2mt+t=15-10m\\ \Leftrightarrow m^2t+2m\left(t+5\right)+t-15=0\)

PT có nghiệm nên \(\Delta'=\left(t+5\right)^2-t\left(t-15\right)\ge0\)

\(\Leftrightarrow10t+25+15t\ge0\Leftrightarrow t\ge-1\)

Vậy \(xy_{min}=-1\Leftrightarrow\dfrac{5\left(2m-3\right)}{\left(m+1\right)^2}=1\Leftrightarrow m^2-8m+16=0\Leftrightarrow m=4\)

Câu 4: \(a^2+b^2=4a+bc+540\)

c đâu ra vậy?

Câu 5:

Thay \(x=3\Leftrightarrow P\left(2\right)+2P\left(2\right)=3^2\Leftrightarrow P\left(2\right)=3\)

Thay \(x=\sqrt{2013}\)

\(\Leftrightarrow P\left(\sqrt{2013}-1\right)+2P\left(2\right)=\left(\sqrt{2013}\right)^2=2013\\ \Leftrightarrow P\left(\sqrt{2013}-1\right)+6=2013\\ \Leftrightarrow P\left(\sqrt{2013}-1\right)=2007\)

- 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 ....

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

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne3\\y\ne1\end{matrix}\right.\)

Đặt `(x)/(x-3)` = a, `(y)/(y-1)` = b

  \(\text{Hệ}\Leftrightarrow\left\{{}\begin{matrix}a+3b=5\\4a-b=7\end{matrix}\right.\\ \Leftrightarrow...\\ \Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x-3}=2\\\dfrac{y}{y-1}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2x-6\\y=y-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=6\\-1=0\left(vô.lí\right)\end{matrix}\right.\)

Vậy hpt vô nghiệm

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

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

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

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