\(x^{671}+y^{671}=1\Rightarrow\left(x^{671}+y^{671}\right)^2=x^{1342}+2.x^{671}.y^{671}+y^{1342}\)\(=1\)

Mà \(x^{1342}+y^{1342}=2\) \(\Rightarrow x^{671}.y^{671}=\dfrac{-1}{2}\)

Mặt khác: \(\left(x^{671}+y^{671}\right)^3=x^{2013}+3x^{671}y^{671}\left(x^{671}+y^{671}\right)+y^{2013}=1\)

Hay \(x^{2013}+y^{2013}-\dfrac{3}{2}.1=1\Rightarrow x^{2013}+y^{2013}=1+\dfrac{3}{2}=\dfrac{5}{2}\)

thiếu đề. (2)

`(x-1)^2>=0`

\(x^{671}=a;\text{ }y^{671}=b\)

Ta có: \(a+b=0,67;\text{ }a^2+b^2=1,34\)

\(\Rightarrow b=0,67-a;\text{ }a^2+\left(0,67-a\right)^2=1,34\)

Giải phương trình theo công thức nghiệm (chính xác) rồi láp vô máy tính giá trị cần tìm.

\(b,\hept{\begin{cases}4\left(x+y\right)=5\left(x-y\right)\\\frac{40}{x+y}+\frac{40}{x-y}=9\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}4\left(x+y\right)^2\left(x-y\right)-5\left(x-y\right)^2\left(x+y\right)=0\\40\left(x-y\right)+40\left(x+y\right)-9\left(x-y\right)\left(x+y\right)=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(x^2-y^2\right)\left[4\left(x+y\right)-5\left(x-y\right)\right]=0\\80x-9\left(x^2-y^2\right)=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(x+y\right)\left(x-y\right)\left(9y-x\right)=0\\9\left(\frac{80}{9}x-x^2+y^2\right)=0\end{cases}}\)

\(\Rightarrow.......\)

rồi sao típ ạ?

Bạn xét phương trình thứ nhất 

bằng cách chia trường hợp rồi tìm

\(DK:\hept{\begin{cases}x>0&y>\frac{2012}{2013}&\end{cases}}\)

HPT

\(\text{ }\Leftrightarrow\hept{\begin{cases}2013\sqrt{2013y-2012}=\frac{2013}{x}\left(1\right)\\y^2+2012=\frac{2013}{x}\left(2\right)\end{cases}}\)

\(\left(1\right),\left(2\right)\Rightarrow y^2-2013\sqrt{2013y-2012}+2012=0\)

\(\Leftrightarrow\left(y^2-1\right)-2013\left(\sqrt{2013y-2012}-1\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(y-1\right)-\frac{2013^2\left(y-1\right)}{\sqrt{2013y-2012}+1}=0\)

\(\Leftrightarrow\left(y-1\right)\left(y+1-\frac{2013^2}{\sqrt{2013y-2012}+1}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=1\\y+1-\frac{2013^2}{\sqrt{2013y-2012}+1}=0\end{cases}}\)

Cai PT thu to ay vo nghiem nhung biet chung minh :)

\(\Rightarrow x=1\)

Vay nghiem cua HPT la \(\left(x;y\right)=\left(1;1\right)\)

\(\hept{\begin{cases}\frac{\sqrt{x^2+xy+y^2}}{|x+y|}=\frac{\sqrt{3}}{2}\left(1\right)\\x^{2012}+y^{2012}=2^{2013}\left(2\right)\end{cases}}\)

\(\left(1\right)< =>2\sqrt{x^2+xy+y^2}=\sqrt{3}|x+y|\)

\(< =>4\left(x^2+xy+y^2\right)=3\left(x+y\right)^2\)

\(< =>4x^2+4xy+4y^2=3x^2+6xy+3y^2\)

\(< =>\left(x-y\right)^2=0\)

\(< =>x=y\)

\(\left(2\right)< =>2x^{2012}=2^{2013}\)

\(< =>x^{2012}=2^{2012}\)

\(< =>\orbr{\begin{cases}x=y=2\\x=y=-2\end{cases}}\)

Vậy (x;y) thuộc (2;2) hoặc (-2;-2)

a) \(\hept{\begin{cases}\left(x-1\right)\left(2x+y\right)=0\\\left(y+1\right)\left(2y-x\right)=0\end{cases}}\)
\(\cdot x=1\Rightarrow\hept{\begin{cases}0=0\\\left(y+1\right)\left(2y-1\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}0=0\\y=-1;y=\frac{1}{2}\end{cases}}\)
\(\cdot y=-1\Rightarrow\hept{\begin{cases}\left(x-1\right)\left(2x-1\right)=0\\0=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1;x=\frac{1}{2}\\0=0\end{cases}}\)
\(\cdot x=2y\Rightarrow\hept{\begin{cases}\left(2y-1\right)5y=0\\0=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=0\Rightarrow x=0\\y=\frac{1}{2}\Rightarrow x=1\end{cases}}\)
\(y=-2x\Rightarrow\hept{\begin{cases}0=0\\\left(1-2x\right)5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\Rightarrow y=-1\\x=0\Rightarrow y=0\end{cases}}\)

b) \(\hept{\begin{cases}x+y=\frac{21}{8}\\\frac{x}{y}+\frac{y}{x}=\frac{37}{6}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\\left(\frac{21}{8}-y\right)^2+y^2=\frac{37}{6}y\left(\frac{21}{8}-y\right)\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\2y^2-\frac{21}{4}y+\frac{441}{64}=-\frac{37}{6}y^2+\frac{259}{16}y\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\1568y^2-4116y+1323=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{8}\\y=\frac{9}{4}\end{cases}}hay\hept{\begin{cases}x=\frac{9}{4}\\y=\frac{3}{8}\end{cases}}\)

c) \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\\\frac{2}{xy}-\frac{1}{z^2}=4\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x-y\right)^2=-4x^2y^2+2xy\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}8x^2y^2-4x^2y-4xy^2+x^2+y^2-2xy+2xy=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4x^2y^2-4x^2y+x^2+4x^2y^2-4xy^2+y^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x\right)^2+\left(2xy-y\right)^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=\frac{-1}{2}\end{cases}}\)
d) \(\hept{\begin{cases}xy+x+y=71\\x^2y+xy^2=880\end{cases}}\). Đặt \(\hept{\begin{cases}x+y=S\\xy=P\end{cases}}\), ta có: \(\hept{\begin{cases}S+P=71\\SP=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P\left(71-P\right)=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P^2-71P+880=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S=16\\P=55\end{cases}}hay\hept{\begin{cases}S=55\\P=16\end{cases}}\)
\(\cdot\hept{\begin{cases}S=16\\P=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=16\\xy=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y\left(16-y\right)=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y^2-16y+55=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=5\\y=11\end{cases}}hay\hept{\begin{cases}x=11\\y=5\end{cases}}\)

\(\cdot\hept{\begin{cases}S=55\\P=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=55\\xy=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y\left(55-y\right)=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y^2-55y+16=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{55-3\sqrt{329}}{2}\\y=\frac{55+3\sqrt{329}}{2}\end{cases}}hay\hept{\begin{cases}x=\frac{55+3\sqrt{329}}{2}\\y=\frac{55-3\sqrt{329}}{2}\end{cases}}\)

e) \(\hept{\begin{cases}x\sqrt{y}+y\sqrt{x}=12\\x\sqrt{x}+y\sqrt{y}=28\end{cases}}\). Đặt \(\hept{\begin{cases}S=\sqrt{x}+\sqrt{y}\\P=\sqrt{xy}\end{cases}}\), ta có \(\hept{\begin{cases}SP=12\\P\left(S^2-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\P\left(\frac{144}{P^2}-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\2P^4+28P^2-144P=0\end{cases}}\)
Tự làm tiếp nhá! Đuối lắm luôn