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

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.

giúp với ạ

Nhờ mn giúp mk nha!

(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

\(5x.\left(x-1\right)=x-1\)

\(5x.\left(x-1\right)-\left(x-1\right)=0\)

\(\left(x-1\right).\left(5x-1\right)=0\)

=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\5x=1\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

\(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\5x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

  Vậy \(x\in\left\{1;\frac{1}{5}\right\}\)

\(2.\left(x+5\right)-x^2-5x=0\)

\(2.\left(x+5\right)-\left(x^2+5x\right)=0\)

\(2.\left(x+5\right)-x.\left(x+5\right)=0\)

\(\left(x+5\right).\left(2-x\right)=0\)

=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

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

MÌnh Bị Thiếu Cho x+y=7