Bài 1:
\(a-c=m^2+n^2-2mn=\left(m-n\right)^2>0\)
\(\Rightarrow a>c\)
\(a-b=m^2+n^2-m^2+n^2=2n^2>0\)
\(\Rightarrow a>b\)
\(a-\left(b+c\right)=m^2+n^2-\left(m^2-n^2+2mn\right)=2n^2-2mn=2n\left(n-m\right)< 0\)
\(\Rightarrow b+c>a\) mà \(a>b,a>c\)
\(\Rightarrow a,b,c\) là độ dài 3 cạnh của 1 tam giác.
Ta có: \(b^2+c^2=\left(m^2-n^2\right)+4m^2n^2=m^4-2m^2n^2+n^4+4m^2n^2=m^4+2m^2n^2+n^4=\left(m^2+n^2\right)^2\)
\(a^2=\left(m^2+n^2\right)^2\)
\(\Rightarrow a^2=b^2+c^2\)
\(\Rightarrow a,b,c\) là độ dài 3 cạnh của tam giác vuông (định lí Py-ta-go đảo).
Bài 2:
a) \(a^2-b^2-c^2+2bc=a^2-\left(b-c\right)^2=\left(a-b+c\right)\left(a+b-c\right)=\left(2m-2b\right)\left(2m-2c\right)=4\left(m-b\right)\left(m-c\right)\left(đpcm\right)\)
We have \(a+b=m^2+n^2+m^2-n^2=2m^2\) and \(c=2mn\)
Because \(m>n>0\Leftrightarrow2m^2>2mn\)
From these, we get \(a+b>c\) (1)
We have \(a+c=m^2+n^2+2mn=\left(m+n\right)^2\) and \(b=m^2-n^2=\left(m+n\right)\left(m-n\right)\)
Because \(m>n>0\), we have \(m+n>m-n\Leftrightarrow\left(m+n\right)^2>\left(m+n\right)\left(m-n\right)\). Therefore, \(a+c>b\) (2)
And, \(b+c=m^2-n^2+2mn\) and \(a=m^2+n^2\)
Because \(m>n\Leftrightarrow2mn>2n^2\Leftrightarrow-n^2+2mn>n^2\)\(\Leftrightarrow m^2-n^2+2mn>m^2+n^2\)
From these, we have \(b+c>a\) (3)
Apparently, \(a=m^2+n^2>0\), \(c=2mn>0\) and \(b=m^2-n^2>0\) with \(m>n>0\)
From (1), (2), (3), we have \(\left\{{}\begin{matrix}a+b>c\\a+c>b\\b+c>a\end{matrix}\right.\). Thus, \(a,b,c\) is the 3 length of the 3 sides of a triangle.
