1. Ta có: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
Theo đề ta có: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab=5^2-4.2=17\)
Vậy \(\left(a-b\right)^2=17\)
2. Ta có: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
Theo đề ta có: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab=6^2+4.16=100\)
\(\Rightarrow\left[{}\begin{matrix}a+b=10\\a+b=-10\end{matrix}\right.\)
Vậy \(a+b=10\) hoặc \(a+b=-10\)
3. \(a^2+b^2+1=ab+a+b\)
\(\Rightarrow2\left(a^2+b^2+1\right)=2\left(ab+a+b\right)\)
\(\Rightarrow2a^2+2b^2+2=2ab+2a+2b\)
\(\Rightarrow2a^2+2b^2+2-2ab-2a-2b=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\) (1)
Vì \(\left(a-b\right)^2\ge0\) \(\forall a;b\)
\(\left(a-1\right)^2\ge0\) \(\forall a\)
\(\left(b-1\right)^2\ge0\) \(\forall b\)
\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) \(\forall a;b\) (2)
Từ (1)(2) \(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-1\right)^2=0\\\left(b-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\Rightarrow a=b=1\)
Vậy.... đpcm
Chúc bạn học tốt ahihi
Bài 1 : \(a+b=5\)
\(\Leftrightarrow\left(a+b\right)^2=25\)
\(\Leftrightarrow a^2+b^2+2ab=25\)
\(\Leftrightarrow a^2+b^2+2.2=25\)
\(\Leftrightarrow a^2+b^2=21\)
\(\Leftrightarrow a^2+b^2-2ab=21-2ab\)
\(\Leftrightarrow\left(a-b\right)^2=21-2.2\)
\(\Leftrightarrow\left(a-b\right)^2=17\)
Bài 2 :
\(a-b=6\)
\(\Leftrightarrow\left(a-b\right)^2=36\)
\(\Leftrightarrow a^2-2ab+b^2=36\)
\(\Leftrightarrow a^2+b^2-2.16=36\)
\(\Leftrightarrow a^2+b^2=36+32=68\)
\(\Leftrightarrow a^2+b^2+2ab=68+2ab\)
\(\Leftrightarrow\left(a+b\right)^2=68+2.16=100\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=10\\a+b=-10\end{matrix}\right.\)
Bài 3 :
\(a^2+b^2+1=ab+a+b\)
\(\Leftrightarrow2\left(a^2+b^2+1\right)=2\left(ab+a+b\right)\)
\(\Leftrightarrow2a^2+2b^2+2=2ab+2a+2b\)
\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(a-1\right)^2\ge0;\left(b-1\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)
Vậy \(a=b=1\)
