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a: Đặt |x-6|=a, |y+1|=b
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a+3b=5\\5a-4b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
=>|x-6|=1 và |y+1|=1
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)
b: Đặt |x+y|=a, |x-y|=b
Theo đề, ta có: \(\left\{{}\begin{matrix}2a-b=19\\3a+2b=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{55}{7}\\b=-\dfrac{23}{7}\left(loại\right)\end{matrix}\right.\)
=>HPTVN
c: Đặt |x+y|=a, |x-y|=b
Theo đề ta có: \(\left\{{}\begin{matrix}4a+3b=8\\3a-5b=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)
=>|x+y|=2 và x=y
=>|2x|=2 và x=y
=>x=y=1 hoặc x=y=-1
1) \(\left\{{}\begin{matrix}x+y=1\\x^3+y^3=x^2+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x^2+y^2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\x^2-xy+y^2-x^2-y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=1\\y=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x^2+y^2=5\\x^4-x^2y^2+y^4=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\\left(x^2+y^2\right)^2-3x^2y^2=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\\left(5\right)^2-3x^2y^2=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\-3x^2y^2=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\x^2y^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\\left(5-y^2\right)y^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\-y^4+5y^2-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\-y^4+5y^2-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\\left[{}\begin{matrix}y^2=1\\y^2=4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2=5-y^2\\y^2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x^2=5-y^2\\y^2=4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2=4\\y^2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x^2=1\\y^2=4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
vậy \(S=\left\{\left(2;1\right),\left(2;-1\right),\left(-2;1\right),\left(-2;-1\right),\left(1;2\right),\left(1;-2\right),\left(-1;2\right),\left(-1;-2\right)\right\}\)
4) \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^5-1=-y^5\\x^9-x^4+y^9-y^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x^5-1=-y^5\\y^5-1=-x^5\end{matrix}\right.\\x^4\left(x^5-1\right)+y^4\left(y^5-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^5+y^5=1\\x^4\left(-y^5\right)+y^4\left(-x^5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^5+y^5=1\\-x^4y^4\left(x+y\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^5+y^5=1\\\left[{}\begin{matrix}x=0\\y=0\\x+y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^5+y^5=1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}x^5+y^5=1\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x^5+y^5=1\\x=-y\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-y\\x^5-x^5=1\end{matrix}\right.\end{matrix}\right.\)
vậy \(S=\left\{\left(1;0\right),\left(0;1\right)\right\}\)
3) ta xét phương trình thứ nhất
\(x-\frac{1}{x}=y-\frac{1}{y}\)
<=>\(x-y-\frac{1}{x}+\frac{1}{y}=0\)
<=>\(x-y-\left(\frac{1}{x}-\frac{1}{y}\right)=0\)
<=>\(x-y-\left(\frac{y-x}{xy}\right)=0\)
<=>\(\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\)
<=>\(x=y\) hoặc xy=-1
Với x=y thay vào phương trình thứ hai ta có
\(2x=x^3+1
\)
<=> \(x^3-2x+1=0\)
<=>\(x^3-x^2+x^2-x-x+1=0\)
<=>\(\left(x-1\right)\left(x^2+x-1\right)=0\)
<=> \(x=1\) hoặc \(x^2+x-1=0\)
\(x^2+x-1=0\) <=> \(x=\frac{-1+\sqrt{5}}{2}\)
hoặc \(x=\frac{-1-\sqrt{5}}{2}\)
Đối với xy=-1 thì y=-1/x thay vào phương trình 2 giải bình thường
a/ \(x^4+y^4=1\Rightarrow\left\{{}\begin{matrix}x^4\le1\\y^4\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|x\right|\le1\\\left|y\right|\le1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^4\ge x^6\\y^4\ge y^6\end{matrix}\right.\) \(\Rightarrow x^6+y^6\le x^4+y^4\le1\)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x^4=x^6\\y^4=y^6\\x^4+y^4=1\end{matrix}\right.\)
\(\Leftrightarrow\left(x;y\right)=\left(1;0\right);\left(0;1\right);\left(-1;0\right);\left(0;-1\right)\)
b/ \(\Rightarrow x^9+y^4=1.\left(x^4+y^4\right)\)
\(\Rightarrow x^9+y^9=\left(x^5+y^5\right)\left(x^4+y^4\right)\)
\(\Rightarrow x^9+y^9=x^9+y^9+x^5y^4+x^4y^5\)
\(\Rightarrow x^4y^4\left(x+y\right)=0\Rightarrow\left[{}\begin{matrix}xy=0\\x=-y\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(0;1\right);\left(1;0\right)\)