x2 + 10x + 26 + y2 + 2y
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(a^2+b^2+2ab+2a+2b+1=\left(a^2+2ab+b^2\right)+\left(2a+2b\right)+1\)
\(=\left(a+b\right)^2+2\left(a+b\right)+1=\left[\left(a+b\right)+1\right]^2=\left(a+b+1\right)^2\)
b) K phân tích dc.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=3(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)=2^32-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5x.\left(x-1\right)=x-1\)
\(5x.\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right).\left(5x-1\right)=0\)
=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\5x=1\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
\(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\5x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(x\in\left\{1;\frac{1}{5}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2.\left(x+5\right)-x^2-5x=0\)
\(2.\left(x+5\right)-\left(x^2+5x\right)=0\)
\(2.\left(x+5\right)-x.\left(x+5\right)=0\)
\(\left(x+5\right).\left(2-x\right)=0\)
=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
2(x+5)-x2-5x=0
2(x+5)-(x2+5x)=0
2(x+5)-x(x+5)=0
(2-x)(x+5)=0
2-x=0 hoặc x+5=0
x=2 hoặc x=-5
hình như bài này hỏi GTNN nhưng bn này k giỏi toán nên viết k đúng
= (x+5)2 + (y+1)2
Vậy GTNN = 0