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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
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 \(S=\left\{\left(1;0\right),\left(0;1\right)\right\}\)